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https://socratic.org/questions/how-do-you-find-the-domain-and-range-of-f-x-1-2-abs-x-2
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# How do you find the domain and range of f(x)=(1/2)abs(x-2)?
$x$ can have any value, $f \left(x\right)$ can only be positive or zero
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2019-11-17 19:54:38
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http://atozmath.com/Default.aspx?q1=Is%20Identity%20matrix%20%5B%5B1%2C0%2C0%5D%2C%5B0%2C1%2C0%5D%2C%5B0%2C0%2C1%5D%5D%60518&do=1
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Home > Matrix Algebra calculators > is Identity Matrix calculator
Solve any problem (step by step solutions) Input table (Matrix, Statistics)
Mode :
SolutionHelp
Solution
Find Is Identity matrix [[1,0,0],[0,1,0],[0,0,1]]
Solution:
Your problem -> Is Identity matrix [[1,0,0],[0,1,0],[0,0,1]]
A square matrix, in which all diagonal elements are unity and all other elements are zero, is called an identity matrix or a unit matrix.
Or
A diagonal matrix, in which all diagonal elements are unity, is called an identity matrix or a unit matrix.
A =
1 0 0 0 1 0 0 0 1
Here, all diagonal elements are unity and all other elements are zero, so it is an identity matrix or a unit matrix.
Solution provided by AtoZmath.com
Any wrong solution, solution improvement, feedback then Submit Here
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2018-12-14 08:11:36
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https://discuss.codechef.com/t/editorial-for-xorney-pelt2019/21596
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# Editorial for XORNEY - PELT2019
Question Code: XORNEY
Prerequisites: Implementation
Problem Setter: tds115
Difficulty: Cakewalk
Explanation:
Given two integer l and r, you have to find if xor from l to r is even or odd. Basically, you have
to find the number of odd integer between l and r. Since xor of odd 1’s is 1 and xor of even
1’s is 0. You can make 2 cases:-
(Let count be the number of odd’s)
If l is even and r is even-> count=(r-l)/2
Else -> count=(r-l)/2+1
If count is even then ans is even
Else odd.
Complexity-O(1) for each query
Author’s solution: https://ideone.com/GUjqC5
2 Likes
Given that this is a simple problem, it probably needs a more basic explanation.
If we have a “running total” that we are xor’ing numbers into, the parity (even/odd) only changes when we xor an odd number into it, because parity only depends on the bottom binary bit.
One odd number makes the running total odd; two odd numbers flip it back to even.
So - as you say - we are determining how many odd numbers there are between L and R inclusive.
If L & R are both even, this is easy: (R-L)/2 gives the number of odd numbers in range (try it!)
If L is odd, we can drop L down to the even number below (L-1) without changing the count of odd numbers in range. And if R is odd, we can increase R to the even number above without changing the count. So we can get to two even numbers to use the above formula.
And we can do those adjustments automatically using modulo results (%), since a%2 is zero for even a. So we can take C = ((R+R%2) - (L-L%2)) /2 for the count of odd numbers in range.
Then the parity of C is the answer, C%2 == 0 is “Even” and C%2 == 1 is “Odd”.
1 Like
/* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
while (t-- > 0)
{
long L = Integer.valueOf(values[0]);
long R = Integer.valueOf(values[1]);
// counting odd numbers
long count = (R - L) / 2;
if ((R % 2 != 0) || (L % 2 != 0))
count = count + 1;
System.out.println((count % 2 == 0) ? "Even" : "Odd");
}
}
}
Status: Runtime Error (NZEC)
What’s wrong?
1 Like
Isn’t it wrong
i mean try these testcases
2
3 3
4 4
the output should be
Even
Even
Odd
Even
which is not right i guess cause any number xored with itself will result in 0 which is considered as “Even”.
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2022-07-05 06:08:47
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http://clay6.com/qa/24562/two-light-waves-whose-equation-are-given-respectively-as-y-1-6-sin-wt-y-2-8
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Want to ask us a question? Click here
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# Two light waves whose equation are given respectively as $y_1 = 6\sin wt; y_2 =8 \sin (wt+ 8)$ super impose to form a interference pattern. The ratio of maximum to minimum intensities of the super imposed wave is
$(a)\;49:1 \\ (b)\;1:49 \\ (c)\;1:7 \\ (d)\;7:1$
Can you answer this question?
## 1 Answer
0 votes
From the two given equation of wave
$a_1 =6 \;a_2= 8$
$\large\frac{I_{max}}{I_{min}}=\large\frac{\bigg[\Large\frac{a_1}{a_2}+1\bigg]^2}{\bigg[\Large\frac{a_1}{a_2} -1\bigg]^2}$
$\qquad=\large\frac{\bigg[\Large\frac{6}{8}+1\bigg]^2}{\bigg[\Large\frac{6}{8} -1\bigg]^2}$
$\qquad= \large\frac{49}{1}$
Hence a is the correct answer.
answered Jan 21, 2014 by
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2017-04-26 17:40:36
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https://gdeq.org/Sokolov_V.V._Algebraic_quantum_Hamiltonians_on_the_plane,_talk_at_The_Mini-Workshop_on_Integrable_Equations,_17_February_2015,_Independent_University_of_Moscow_(abstract)
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# Sokolov V.V. Algebraic quantum Hamiltonians on the plane, talk at The Mini-Workshop on Integrable Equations, 17 February 2015, Independent University of Moscow (abstract)
Title: Algebraic quantum Hamiltonians on the plane
Abstract:
In is known that many of quantum Calogero-Moser type Hamiltonians admit a change of variables bringing them to differential operators ${\displaystyle P}$ of second order with polynomial coefficients.
It turns out that in all known examples:
${\displaystyle \mathbf {1} }$ : ${\displaystyle P}$ preserves some finite-dimensional polynomial vector space ${\displaystyle V}$.
The set of all differential operators with polynomial coefficients that preserve a fixed finite-dimensional polynomial vector space ${\displaystyle V}$ forms an associative algebra ${\displaystyle A}$. In the most interesting case the vector space ${\displaystyle V}$ coincides with the space of all polynomials of degrees ${\displaystyle \leq k}$ for some ${\displaystyle k}$. For such ${\displaystyle V}$ the algebra ${\displaystyle A}$ is the universal enveloping algebra ${\displaystyle \operatorname {sl} (n+1)}$, where ${\displaystyle n}$ is the number of independent variables.
It is clear that if a differential operator satisfies Property ${\displaystyle \mathbf {1} }$, we can find several eigenvalues and corresponding polynomial eigenvectors in an explicit algebraic form.
For the elliptic Calogero type models the flat metric ${\displaystyle g}$ related to the symbol of ${\displaystyle P}$ depends on the elliptic parameter. One of the reasons why such a metric could be interesting in itself is that families of contravariant metrics with linear dependence on a parameter are closely related to the Frobenius manifolds.
Slides: Sokolov V. Algebraic quantum Hamiltonians on the plane (presentation at The Mini-Workshop on Integrable Equations, 17 February 2015, Independent University of Moscow).pdf
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2021-06-25 12:33:13
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https://www.bio-physics.at/wiki/index.php?title=Regulating_vs._Steering
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# Regulating vs. Steering
### The difference between regulating and steering
Control theory is usually deals with the control of certain parameters of a system. A simple example is thermoregulation, in which the controlled parameter is temperature. First you set a certain temperature value at the thermostat as Reference point, at which the control system should keep the temperature. The system measures the current room temperature via a Sensor and compares it to the reference temperature. If a deviation from the reference temperature is measured, the error signal results in the control action of the Controller, in this example a heating device, that is turned up or down, according to the error signal. If it is to cold the heater is tured up, if it is to hot the heater is turned down. In this manner the controlled System is kept at a constant temperature. Control systems of this kind are called feedback control systems, because the output (temperature) is fed back to the input (sensor), forming a closed loop, that is why they are also called closed-loop systems.
temperature forecast for a period of 12 days. red curve: mean temperature; dark blue: deviations from mean
The idea of regulation bears a fundamental extension of the thoughts of early physics. In physics the primary goal was to examine a system to then predict its behavior. Knowledge about a system is regarded as complete, if all observations can be predicted. Now, the fundamental difference in control systems is that the system can be an Open System, that is a system that can have inputs or boundary conditions that are not known in advance and still the system state can be known, since it is regulated. Considering the familiar example of thermoregulation, this means, that the early physics approach might would have been to predict the weather, to then implement the weather time course in a controller, which is then able to keep the room temperature constant. Due to the nonlinear behavior of the weather, it is not possible to predict the weather for arbitrary times, since small deviations in the initial conditions can cause large differences in the calculated behavior of the system (see article Liapunov Exponent).
It is simply not possible to predict the weather for long periods of time. The graph on the right shows, that after twelve days the temperature can be somewhere in the range between $-18°C$ and $+24°C$.
In fact, the difference between the first method (feedback control - closed loop control) and the second method (called open-loop-control) described, is that the feedback systems constantly integrate information - they regulate, while the open-loop systems steer. Another example of steering, a more standard one, is to program a robot to dance, like this one, where you program the motors to move in a predefined way. It is evident, that if the steering is not 'intelligent', lets say you program the robot to lean back too much, he will fall. Or for a more complex situation, for instance the wind blows, you have to know, how hard the wind blows to keep the robot from falling, by doing the right action. Intelligence in this respect means knowledge about all forces acting on the system. Since in real life most of the things happening are not predictable, feedback control provides the 'intelligent' solution without prediction.
In simple control operations like the example of the thermostat shown above, the task of the controller is to keep some parameter constant. If we consider an organism as a biological system, that evolved through evolution, to what extent are such control operations manifested in the organisms? Some examples of these mechanisms are the control of body temperature and the eyes pupil-control-system, which have both been worked out in detail by John H. Milsum in his Book 'Biological Control Systems Analysis', that was already published in 1966. All examples shown so far, whether technical or biological have in common, that the aim of control is known. In technical control operations the goal is given by the designer. But how did the biological systems evolve to control certain parameters? Evolution is thought to optimise the response of the organism in a certain environment. But what is the goal of the response? What gets optimised? In certain situations the growth rate of an organism has prooved to be maximised (see this article). However, since evolution is an open process with no predefined goal, no gerneral statement about a universal goal can be made. Like Shakespeare said, "To Be or not to Be: That is the question"! So that the only thing we can assume to be intrinsic to all existing things is, that its goal is existence, otherwise it would not exist. It means everything that exists does something that makes it exist. But this statement is fairly useless since it has no predictive power. The question then is, what it does, that makes it exist.
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2021-05-18 17:46:58
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https://eccc.weizmann.ac.il/report/2022/097/
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Under the auspices of the Computational Complexity Foundation (CCF)
REPORTS > DETAIL:
### Paper:
TR22-097 | 3rd July 2022 17:51
#### Unstructured Hardness to Average-Case Randomness
TR22-097
Authors: Lijie Chen, Ron D. Rothblum, Roei Tell
Publication: 8th July 2022 23:29
Keywords:
Abstract:
The leading technical approach in uniform hardness-to-randomness in the last two decades faced several well-known barriers that caused results to rely on overly strong hardness assumptions, and yet still yield suboptimal conclusions.
In this work we show uniform hardness-to-randomness results that *simultaneously break through all of the known barriers*. Specifically, consider any one of the following three assumptions:
1. For some $\epsilon>0$ there exists a function $f$ computable by uniform circuits of size $2^{O(n)}$ and depth $2^{o(n)}$ such that $f$ is hard for probabilistic time $2^{\epsilon\cdot n}$.
2. For every $c\in\mathbb{N}$ there exists a function $f$ computable by logspace-uniform circuits of polynomial size and depth $n^2$ such that every probabilistic algorithm running in time $n^{c}$ fails to compute $f$ on a $(1/n)$-fraction of the inputs.
2. For every $c\in\mathbb{N}$ there exists a logspace-uniform family of arithmetic formulas of degree $n^2$ over a field of size $\mathrm{poly}(n)$ such that no algorithm running in probabilistic time $n^{c}$ can evaluate the family on a worst-case input.
Assuming any of these hypotheses, where the hardness is for every sufficiently large input length $n\in\mathbb{N}$, we deduce that $\mathcal{RP}$ can be derandomized in *polynomial time and on *all input lengths*, on average. Furthermore, under the first assumption we also show that $\mathcal{BPP}$ can be derandomized in polynomial time, on average and on all input lengths, with logarithmically many advice bits.
On the way to these results we also resolve two related open problems. First, we obtain an *optimal worst-case to average-case reduction* for computing problems in linear space by uniform probabilistic algorithms; this result builds on a new instance checker based on the doubly efficient proof system of Goldwasser, Kalai, and Rothblum (J. ACM, 2015). Secondly, we resolve the main open problem in the work of Carmosino, Impagliazzo and Sabin (ICALP 2018), by deducing derandomization from weak and general fine-grained hardness hypotheses.
ISSN 1433-8092 | Imprint
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2022-09-28 13:43:17
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https://www.coursehero.com/file/p51mqb8/3-Coin-Toss-Probability-Distribution-of-Heads-Probability-Binomial-Distribution/
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# 3 coin toss probability distribution of heads
• Test Prep
• 2
• 100% (3) 3 out of 3 people found this document helpful
This preview shows page 1 - 2 out of 2 pages.
3 Coin Toss Probability Distribution # of Heads Probability Binomial Distribution – used when probability problem can be reduced to 2 possible independent outcomes, fixed # of trials, probability of success remains the same for each trial Notation P(s) – probability of success P(f) – probability of failure p – numerical probability of success q – numerical probability of failure n – number of trials X – number of successes P(S)=p, P(F)=q=1–p P(X) = n C x × p x × q n-x mean=μ=np variance=σ 2 =npq standard deviation=σ=√npq
Info About Decks of Cards 52 total cards, 4 suits – hearts (red), diamonds (red), clubs (black), spades (black), each with 13 cards Cards in each suit: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King Normal Distribution: Bell-shaped Unimodal Med=Mode=Mean total area under curve=100% Follows Empirical Rule : for a normal distribution , nearly all of the data will fall within three standard deviations of the mean . 68% of data falls w/in 1 standard deviation from the mean 95% fall w/in 2 99.7% fall w/in 3 Chapter 6: Central Limit Theorem Central Limit Theorem – As the sample size increases, the shape of the distribution of the sample means taken with replacement from a population with mean μ and standard deviation σ will approach a normal distribution. Standard deviation of sample means is called the standard error of the mean. Used for groups, e.g. Find the probability that 100 women… *If the sample is greater than 30, don’t need to test for normality. Examples
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2021-09-19 23:06:22
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https://study.com/academy/answer/suppose-that-p-of-x-is-continuous-at-all-real-numbers-and-satisfies-the-following-equations-integral-from-2-to-5-of-p-of-x-with-respect-to-x-equals-4-integral-from-2-to-10-of-p-of-x-with-respect-to.html
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# Suppose that p of x is continuous at all real numbers and satisfies the following equations....
## Question:
Suppose that {eq}p(x) {/eq} is continuous at all real numbers and satisfies the following equations.
{eq}\bullet {\displaystyle \int_2^5} p(x) dx =4 {/eq}
{eq}\bullet {\displaystyle \int_2^{10}} p(x) dx =13 {/eq}
{eq}\bullet {\displaystyle \int_{10}^{25}} p(x) dx =61 {/eq}
{eq}\bullet {\displaystyle \int_{20}^{25}} p(x) dx =36 {/eq}
What is the value of {eq}{\displaystyle \int_5^{20}} (3p(x)-4) dx {/eq}?
## Integration:
Integration is defined as area under the curve. Properties of integration:
{eq}1. \int (f(x) \pm g(x)) dx = \int f(x) dx \pm \int g(x) dx \\ 2. \int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx \ \ \ \ a< c < b\\ {/eq}
{eq}\int_{5}^{10} p(x) = \int_{2}^{10} p(x) - \int_{2}^{5} p(x)\\ \int_{5}^{10} p(x) = 13 -4 = 9\\ \int_{10}^{20} p(x)= \int_{10}^{25} p(x) - \int_{20}^{25} p(x)\\ \int_{10}^{20} p(x)= 61 - 36 = 25\\ {\displaystyle \int_5^{20}} p(x) = \int_{5}^{10} p(x)+ \int_{10}^{20} p(x)\\ {\displaystyle \int_5^{20}} p(x) (3p(x)-4) dx = 3 {\displaystyle \int_5^{20}} p(x) -4\\ {\displaystyle \int_5^{20}} p(x) (3p(x)-4) dx = 3 (\int_{5}^{10} p(x)+ \int_{10}^{20} ) -4\\ {\displaystyle \int_5^{20}} p(x) (3p(x)-4) dx = 3 cdot (9 + 25) -4 = 102-4 = 98 {/eq}
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2019-09-15 05:29:47
|
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|
http://www.math.uri.edu/~dobrush/mth243/sage/ch1206.html
|
Department of Mathematics
MTH243 (Calculus for Functions of Several Variables) SAGE. Chapter 12: Functions of Several Variables Vladimir A. Dobrushkin,Lippitt Hall 202C, 874-5095,dobrush@uri.edu In this course we will use Sage computer algebra system (CAS), which is a free software. The Sage projects are created to help you learn new concepts. Sage is very useful in visualizing graphs and surfaces in three dimensions. Matlab (commercial software) is also available at engineeering labs. Its free version is called Octave. The university has a license for computer algebra system Mathematica, so it is free to use for its students. A student can also use free CASs: SymPy (based on Python), or Maxima.
## Section 12.6. Limits and Continuity
Example 1. A particular case of ellipsoid is a ball:
ContourPlot3D[ 1 == x^2 + y^2 + z^2, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}]
or
clc
clear
figure
fi0=0;
fi1=360;
R=1;
R0=0;
R1=1;
M=30;
dfi=(fi1-fi0)/M;
dR=(R1-R0)/M;
fi=[fi0:dfi:fi1];
aa=pi/180;
theta0=0;
theta1=360;
dtheta=(theta1-theta0)/M;
theta=[theta0:dtheta:theta1];
t=0;
tt=-1;
for j=1:M/2+1
tt=tt+1;
t=tt;
for i=1:M+1
t=t+1;
b(t)=t;
x(i)=R*sin(aa*fi(i))*cos(aa*theta(j));
y(i)=R*sin(aa*fi(i))*sin(aa*theta(j));
z(i)=R*cos(aa*fi(i));
%pause
v(1:6,b(t))=[ x(i) y(i) z(i) fi(i) theta(j) b(t)];
end
plot3(x,y,z)
axis equal
grid on
hold on
end
title('BONUS 12.69')
t=0;
tt=-1;
for j=1:M/2+1
tt=tt+1;
t=tt;
end
for i=1:M+1
t=t+1;
bb(t)=t;
xx(i)=R*sin(aa*fi(j))*cos(aa*theta(i));
yy(i)=R*sin(aa*fi(j))*sin(aa*theta(i));
zz(i)=R*cos(aa*fi(j));
%pause
ang=45;
RR = rotx(ang);
vv = [xx(i);yy(i);zz(i)];
%pause
yyy = RR*vv;
vvv(1:6,bb(t))=[ yyy(1) yyy(2) yyy(3) fi(j) theta(i) bb(t)];
plot3(xx,yy,zz)
plot3(vvv(1,bb(t)),vvv(2,bb(t)),vvv(3,bb(t)),'r*')
axis equal
grid on
hold on
end
|
2017-11-20 13:18:45
|
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|
http://mathhelpforum.com/math-challenge-problems/41113-clocks-time-print.html
|
# Clocks - Time
• Jun 9th 2008, 11:55 AM
MathLearner
Clocks - Time
Clock A,B and Cstrikes every hour. B slows down and take 2 minutes longer than A per hour while C becomes faster and takes a minute less than A per hour. If they strike together at 12 midnight, when will they strike together again...?
ans: 11 am
• Jun 10th 2008, 12:16 PM
Soroban
Hello, MathLearner!
Quote:
Clocks $A, B\text{ and }C$ strike every hour.
$B$ slows down and take 2 minutes longer than $A$ per hour
while $C$ becomes faster and takes a minute less than $A$ per hour.
If they strike together at 12 midnight, when will they strike together again?
Answer: 11 am . . . . I don't agree!
Clock A strikes every 60 minutes.
Clock B strikes every 62 minutes.
Clock C strikes every 59 minutes.
The LCM of 60, 62, and 59 is: .109,740
$109,\!740\text{ minutes} \:=\:1829\text{ hours} \:=\:76\text{ days, }{\color{blue}5\text{ hours}}$
$\text{They will strike together at }{\color{red}5\text{ am}}\text{ (of the 77th day).}$
• Jun 11th 2008, 12:55 AM
MathLearner
I am not sure about this... In the paper i was solving, the right option given was for 11 am.. there was no option as 5 am...
|
2017-08-23 12:14:40
|
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|
http://mathhelpforum.com/geometry/143908-ratios-area-rectangles.html
|
# Math Help - Ratios of Area in Rectangles
1. ## Ratios of Area in Rectangles
These two problems are giving me a lot of trouble. If any one could give me some help in the right direction I'd really appreciate it.
1 In rectangle ABCD, point E is on side AB so that AE = 10 and EB = 5. What fraction of the area of the rectangle is inside triangle AEC?
2 M and N are the midpoints of consecutive sides of a square ABCD with vertex A in between M and N. What is the ratio of the area of triangle AMN to the area of the complete square.
Thanks again.
2. Originally Posted by lyyy94
These two problems are giving me a lot of trouble. If any one could give me some help in the right direction I'd really appreciate it.
1 In rectangle ABCD, point E is on side AB so that AE = 10 and EB = 5. What fraction of the area of the rectangle is inside triangle AEC?
2 M and N are the midpoints of consecutive sides of a square ABCD with vertex A in between M and N. What is the ratio of the area of triangle AMN to the area of the complete square.
Thanks again.
1) Area of ABCD = AB*BC
Area of AEC = 1/2*AE*BC
3. Hello, lyyy94!
1. In rectangle $ABCD$, point $E$ is on side $AB$ so that: . $AE = 10,\;\;EB = 5$
What fraction of the area of the rectangle is inside triangle $AEC$?
Code:
: 10 E 5 :
A o - - - - - - - o - - - o B
| *:::::::::::::* |
| *:::::::::::* |
| *:::::::::* |
x | *:::::::* | x
| *:::::* |
| *:::* |
| *:*|
D o - - - - - - - - - - - o C
15
The length of the rectangle is 15.
The width of the rectangle is $x.$
The area of $\Delta AEC\:=\:\tfrac{1}{2}(10)(x) \:=\:5x$
The area of rectangle $ABCD \:=\:15x$
The fraction (ratio) is: . $\frac{5x}{15x} \;=\;\frac{1}{3}$
2. $M$ and $N$ are midpoints of sides $DA$ and $AB$ of square $ABCD.$
What is the ratio of the area of $\Delta AMN$ to the area of the square?
Make a sketch and the answer is obvious . . .
Code:
N
A o - - - o - - - o B
|:::::* | * |
|:::* | * |
|:* | * |
M o - - - + - - - *
| * | * |
| * | * |
| * | * |
D o - - - * - - - o C
|
2015-11-29 20:25:18
|
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|
https://forum.math.toronto.edu/index.php?PHPSESSID=0f6jdnc8eanf2pk5uf3ta0o7l0&action=printpage;topic=2323.0
|
Toronto Math Forum
MAT334--2020S => MAT334--Lectures & Home Assignments => Chapter 2 => Topic started by: Yan Zhou on February 10, 2020, 04:42:13 PM
Title: 2.2 home assignment question 18
Post by: Yan Zhou on February 10, 2020, 04:42:13 PM
Find the closed form for the given power series.
$$\sum_{n=2}^{\infty}n(n-1)z^{n}$$
hint: divide by $z^{2}$
I tried the hint but still have no idea.
Thanks in advance.
Title: Re: 2.2 home assignment question 18
Post by: Victor Ivrii on February 11, 2020, 07:38:13 AM
Please correct what you typed
Title: Re: 2.2 home assignment question 18
Post by: Yan Zhou on February 12, 2020, 12:36:51 PM
Yes, I just find out that it is different from textbook, and I know how to do it now.
By the way, there is a typo in question 16 which should be $$\sum_{n=1}^{\infty} n(z-1)^{n-1}$$ instead of $$\sum_{n=1}^{\infty} (z-1)^{n-1}$$
In section 2.3,
question 5 should be $$\int_{0}^{2\pi} \frac{d\theta}{2+cos\theta}$$ instead of $1+cos\theta$
question 8 should be $$\int_{0}^{\pi}\frac{d\theta}{1+(sin\theta)^2}$$ the range is from 0 to $\pi$ instead of $2\pi$
question 9 should be "joining $1-i$ to $1+i$".
Title: Re: 2.2 home assignment question 18
Post by: Victor Ivrii on February 12, 2020, 03:12:59 PM
Fixed. Thanks!
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2022-06-30 13:21:35
|
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|
http://www.probabilistic-numerics.org/en/v0.1.3/public_api/diffeq.html
|
# probnum.diffeq¶
Differential Equations.
This package defines common dynamical models and probabilistic solvers for differential equations.
## Functions¶
logistic(timespan, initrv[, params]) Initial value problem (IVP) based on the logistic ODE. fitzhughnagumo(timespan, initrv[, params]) Initial value problem (IVP) based on the FitzHugh-Nagumo model. lotkavolterra(timespan, initrv[, params]) Initial value problem (IVP) based on the Lotka-Volterra model. seir(timespan, initrv[, params]) Initial value problem (IVP) based on the SEIR model. rigidbody(timespan, initrv) Initial value problem (IVP) for rigid body dynamics without external forces vanderpol(timespan, initrv[, params]) Initial value problem (IVP) based on the Van der Pol Oscillator. threebody(timespan, initrv[, params]) Initial value problem (IVP) based on a three-body problem. probsolve_ivp(ivp[, method, which_prior, …]) Solve initial value problem with Gaussian filtering and smoothing. ivp2ekf0(ivp, prior, evlvar) Computes measurement model and initial distribution for KF based on IVP and prior. ivp2ekf1(ivp, prior, evlvar) Computes measurement model and initial distribution for EKF based on IVP and prior. ivp2ukf(ivp, prior, evlvar) Computes measurement model and initial distribution for EKF based on IVP and prior.
## Classes¶
ODE(timespan, rhs[, jac, hess, sol]) Ordinary differential equations. IVP(timespan, initrv, rhs[, jac, hess, sol]) Initial value problems (IVP). ODESolver(ivp, order) Interface for ODESolver. GaussianIVPFilter(ivp, gaussfilt, with_smoothing) ODE solver that behaves like a Gaussian filter. StepRule(firststep) (Adaptive) step size rules for ODE solvers. ConstantSteps(stepsize) Constant step size rule for ODE solvers. AdaptiveSteps(firststep, atol, rtol[, …]) Adaptive step size selection using proportional control. ODESolution(times, rvs, solver) Gaussian IVP filtering solution of an ODE problem.
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2021-08-06 04:38:33
|
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https://brilliant.org/problems/crazy-pair-of-dice/
|
# Crazy Pair of Dice...
Lets roll a pair of standard 6-sided dice first.
There is one way of obtaining a 2, two ways of obtaining a 3, and so on, up to one way of obtaining a 12.
Now, let there be a pair of another 6-sided dice A and B. (Both non-standard)
A and B satisfy these properties:
• Each face has at least one dot.
• The number of ways of obtaining each sum is the same as for the standard dice.
Let the number of dots on faces of A and B be represented by $$a_i$$ and $$b_i$$ respectively. $$(i\in \{1,2,3,4,5,6\})$$
Calculate $$\displaystyle\sum_{i=1}^{6} (a_i^2+b_i^2)+\left (\displaystyle\sum_{i=1}^6 a_i\right )\left (\displaystyle\sum_{i=1}^6 b_i\right )$$.
×
|
2017-03-25 09:45:58
|
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http://cms.math.ca/cmb/kw/building
|
Search results
Search: All articles in the CMB digital archive with keyword building
Expand all Collapse all Results 1 - 2 of 2
1. CMB 2006 (vol 49 pp. 321)
Balser, Andreas
Polygons with Prescribed Gauss Map in Hadamard Spaces and Euclidean Buildings We show that given a stable weighted configuration on the asymptotic boundary of a locally compact Hadamard space, there is a polygon with Gauss map prescribed by the given weighted configuration. Moreover, the same result holds for semistable configurations on arbitrary Euclidean buildings. Keywords:Euclidean buildings, Hadamard spaces, polygonsCategory:53C20
2. CMB 2001 (vol 44 pp. 385)
Ballantine, Cristina M.
A Hypergraph with Commuting Partial Laplacians Let \$F\$ be a totally real number field and let \$\GL_{n}\$ be the general linear group of rank \$n\$ over \$F\$. Let \$\mathfrak{p}\$ be a prime ideal of \$F\$ and \$F_{\mathfrak{p}}\$ the completion of \$F\$ with respect to the valuation induced by \$\mathfrak{p}\$. We will consider a finite quotient of the affine building of the group \$\GL_{n}\$ over the field \$F_{\mathfrak{p}}\$. We will view this object as a hypergraph and find a set of commuting operators whose sum will be the usual adjacency operator of the graph underlying the hypergraph. Keywords:Hecke operators, buildingsCategories:11F25, 20F32
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2013-12-06 05:52:29
|
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http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node65.html
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# 8.5 Rankine Power Cycles
A schematic of the components of a Rankine cycle is shown in Figure 8.11. The cycle is shown on - , - , and - coordinates in Figure 8.12. The processes in the Rankine cycle are as follows:
1. : Cold liquid at initial temperature is pressurized reversibly to a high pressure by a pump. In this process, the volume changes slightly.
2. : Reversible constant pressure heating in a boiler to temperature .
3. : Heat added at constant temperature (constant pressure), with transition of liquid to vapor.
4. : Isentropic expansion through a turbine. The quality decreases from unity at point to .
5. : Liquid-vapor mixture condensed at temperature by extracting heat.
[ - coordinates] [ - coordinates] [ - coordinates]
In the Rankine cycle, the mean temperature at which heat is supplied is less than the maximum temperature, , so that the efficiency is less than that of a Carnot cycle working between the same maximum and minimum temperatures. The heat absorption takes place at constant pressure over , but only the part is isothermal. The heat rejected occurs over ; this is at both constant temperature and pressure.
To examine the efficiency of the Rankine cycle, we define a mean effective temperature, , in terms of the heat exchanged and the entropy differences:
The thermal efficiency of the cycle is
The compression and expansion processes are isentropic, so the entropy differences are related by
The thermal efficiency can be written in terms of the mean effective temperatures as
For the Rankine cycle, , . From this equation we see not only the reason that the cycle efficiency is less than that of a Carnot cycle, but the direction to move in terms of cycle design (increased ) if we wish to increase the efficiency.
There are several features that should be noted about Figure 8.12 and the Rankine cycle in general:
1. The - and the - diagrams are not similar in shape, as they were with the perfect gas with constant specific heats. The slope of a constant pressure reversible heat addition line is, as derived in Chapter 6,
In the two-phase region, constant pressure means also constant temperature, so the slope of the constant pressure heat addition line is constant and the line is straight.
2. The effect of irreversibilities is represented by the dashed line from to . Irreversible behavior during the expansion results in a value of entropy at the end state of the expansion that is higher than . The enthalpy at the end of the expansion (the turbine exit) is thus higher for the irreversible process than for the reversible process, and, as seen for the Brayton cycle, the turbine work is thus lower in the irreversible case.
3. The Rankine cycle is less efficient than the Carnot cycle for given maximum and minimum temperatures, but, as said earlier, it is more effective as a practical power production device.
Muddy Points
Where does degrees Rankine come from? Related to Rankine cycles? (MP 8.9)
UnifiedTP
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2014-08-01 16:12:30
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http://mathhelpforum.com/calculus/127504-3-times-differentiable-but-not-3-times-continuously-differentiable.html
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# Math Help - 3 times differentiable, but not 3 times continuously differentiable?
1. ## 3 times differentiable, but not 3 times continuously differentiable?
Does anyone know an example of a function that is 3 times differentiable, but not 3 times continuously differentiable?
2. Originally Posted by paupsers
Does anyone know an example of a function that is 3 times differentiable, but not 3 times continuously differentiable?
Try $f(x) = x^6\sin(1/x)$ (with f(0) = 0). The third derivative will exist everywhere but be discontinuous at the origin.
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2015-01-25 11:22:42
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https://studydaddy.com/question/acc-230-week-3-discussion-questions
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QUESTION
# ACC 230 Week 3 Discussion Questions
This work of ACC 230 Week 3 Discussion Questions includes:
DQ1:
Post your answer to Problem 3.5 on p. 109 (Ch. 3), and then respond to the following
question: How can the information contained within the stockholder equity statement be
used for management and investor decision making? Provide specific examples of
situations in which the stockholder equity information might be used.
DQ2:
- Provide an example from the text or the Internet that demonstrates a situation in which a
company’s net profits looked good in the statements, but the gross or operating profits
presented a different picture. Discuss how this might have occurred.
- Respond to the question addressed in Problem 3.6 on p. 109 (Ch. 3): “Why is the
bottom-line figure, net income, not necessarily a good indicator of a firm’s financial
success?” (Hint: Look for indicators like liquidity or solvency to answer this discussion
question.)
• @
• 1 order completed
Tutor has posted answer for $5.19. See answer's preview$5.19
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2018-04-26 13:43:28
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http://math.stackexchange.com/questions/19917/tangent-bundle-on-s3
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# Tangent Bundle on S^3
how to show T(S^3) isomorphic to S^3 cross R^3? so can I say it for every odd dimension?I have shown it for n=1
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You can't make a proof which works for every odd $n$, since it's only true for $n=1$, $n=3$ and $n=7$. en.wikipedia.org/wiki/Parallelizable_manifold – Hans Lundmark Feb 1 '11 at 17:38
This is a fact about Lie groups. You can use the group structure to translate a local trivialization around the manifold. However, it also works for $H$-spaces such as $S^7$. – Sean Tilson Feb 3 '11 at 0:44
You want to show that the tangent bundle T(S^3) is a trivial bundle. It is a (not so hard) theorem that a vector bundle being trivial is equivalent to the existence of a (global) basis of sections. A section of the tangent bundle is another word for a vector field.
For S^1, considered as the unit circle in the complex plane, you probably have considered the vector field $z\mapsto iz$, which forms a basis. For S^3, you can do something similar, just like Mariano has described: now you consider S^3 as the unit sphere in the quaternions, and look at the vector fields $V_i:z\mapsto iz$ $V_j:z\mapsto jz$ $V_k:z\mapsto kz$
[Lack of reputation prevents me from being able to place this as a comment. I wanted to point out a similarity between the S^1 and S^3 case.]
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Identify the sphere with the set $S\subset\mathbb H$ of quaternions of norm $1$. Pick an ordered basis $(u_1,u_2,u_3)$ of the tangent space $T_1S$ to $S$ at the point $1$, and now consider the vector fields $$X_i:p\in S \mapsto pu_i\in T_pS,$$ where on the right hand side $pu_i$ is the quaternion product of the element $p$ with the quaternion $u_i$ (recall that one can identify the tangent space at $T_1S$ with a subspace of $\mathbb H$) You need to check that this makes sense: in particular, that $pu_i$ is, in fact, in the tangent space $T_pS$ to $S$ at $p$.
This construction gives you three non-zero vector fields $X_1$, $X_2$ and $X_3$. Check that at each point they are a basis. Voilà.
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It does not work for any odd dimension, if I recall correctly it will only work for 1,3 and 7 which correspond to real-division algebras...
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Actually, doesn't every S^(2n-1) admit the nowhere-zero tangent vector field:
(-x2,x1,-x4,x3,..,-x2n )? (using that S^(2n-1)={ x in R^2n , ||x||=1} )
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Yes. But it doesn't admit a basis of nowhere zero vector fields. I.e. you can't find 2n-1 examples like this, which are everywhere independent. – David Speyer Feb 3 '11 at 12:03
are there any S^(2n-1) 's that have trivial bundles other than for S^3,S^5 and S^7 as Lie groups? Is there some general result in this area? – user6600 Feb 4 '11 at 8:44
Did you miss Hans Lundmark's comment? The only parallelizable S^n's are those with n=1,3,7. – wildildildlife Feb 4 '11 at 14:19
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2016-06-29 11:16:34
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https://support.bioconductor.org/p/90431/
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The descriptions of score methods in DiffBind
1
0
Entering edit mode
Gary ▴ 20
@gary-7967
Last seen 2.6 years ago
Hi,
The descriptions of score methods in DiffBind manual are trimmed and cannot be understood in its pdf file (fig1). Could you provide them for me to learn? Many thanks. By the way, the default is DBA_SCORE_TMM_MINUS_FULL?
Best,
Gary
fig1
DiffBind dba.count score manual PDF • 663 views
1
Entering edit mode
Rory Stark ★ 4.1k
@rory-stark-5741
Last seen 16 days ago
CRUK, Cambridge, UK
You are correct about the default score, however this is only used for plots, and only when the entire experiment if plotted. When data form a speicic contrast is used, the normalised counts from that comparison are plotted and/or reported.
The best way to get all the text is within R:
> library(DiffBind)
> ?dba.count
Cheers-
Rory
0
Entering edit mode
Thank you so much for your help.
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2021-10-20 21:07:43
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http://www.computmath.com/jssx/CN/10.12286/jssx.2020.4.385
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• 青年评述 •
### 复合凸优化的快速邻近点算法
1. 复旦大学大数据学院, 上海数学中心, 上海 200433
• 收稿日期:2020-09-28 出版日期:2020-11-15 发布日期:2020-11-15
• 作者简介:郦旭东,复旦大学大数据学院青年研究员,上海数学中心青年研究员.2010年本科毕业于中国科学技术大学数学系,2015年在新加坡国立大学数学系获博士学位.博士毕业后曾在新加坡国立大学数学系与美国普林斯顿大学运筹与金融工程系任博士后研究员,2018年9月入职复旦大学,于2019年获得由国际数学优化协会(Mathematical Optimization Society)所颁发的连续优化青年学者最佳论文奖,2020年入选第五届中国科协青年人才托举工程,现任期刊《Mathematical Programming Computation》编委.
• 基金资助:
国家自然科学基金（11901107），中国科协青年人才托举工程（2019QNRC001），上海市扬帆计划（19YF1402600），上海市科委项目（19511120700）资助.
Li Xudong. EFFICIENT PROXIMAL POINT ALGORITHM FOR CONVEX COMPOSITE OPTIMIZATION[J]. Mathematica Numerica Sinica, 2020, 42(4): 385-404.
### EFFICIENT PROXIMAL POINT ALGORITHM FOR CONVEX COMPOSITE OPTIMIZATION
Li Xudong
1. School of Data Science, and Shanghai Center for Mathematical Sciences, Fudan University, Shanghai 200433, China
• Received:2020-09-28 Online:2020-11-15 Published:2020-11-15
In the Big Data era, with the advent of convenient automated data collection technologies, large-scale composite convex optimization problems are ubiquitous in many applications, such as massive data analysis, machine and statistical learning, image and signal processing. In this paper, we review a class of efficient proximal point algorithms for solving the large-scale composite convex optimization problems. Under the easy-to-implement stopping criteria and mild calmness conditions, we show the proximal point algorithm enjoys global and local asymptotic superlinear convergence. Meanwhile, based on the duality theory, we propose an efficient semismooth Newton method for handling the subproblems in the proximal point algorithm. Lastly, to further accelerate the proximal point algorithm, we fully exploit the nonsmooth second order information induced by the nonsmooth regularizer in the problem to achieve a dramatic reduction of the computational costs of solving the involved semismooth Newton linear systems.
MR(2010)主题分类:
()
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[1] 申远, 李倩倩, 吴坚. 一种基于邻近点算法的变步长原始-对偶算法[J]. 计算数学, 2018, 40(1): 85-95. [2] 徐海文, 孙黎明. 一类凸优化的加速混合下降算法[J]. 计算数学, 2017, 39(2): 200-212. [3] 徐海文. 一类凸优化的混合下降算法[J]. 计算数学, 2012, 34(1): 93-102.
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2021-08-01 11:54:39
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https://mathswithdavid.com/12-january/
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# 12 January
👆🏼Calculate 203
👇🏼Triangles A and B are mathematically similar. The area of triangle A is 21cm2. Calculate the area of triangle B:
👉🏼Dasha’s bedroom wall measures $6 \frac{2}{3} m$ by $3 \frac{1}{7} m$. She wants to paint it. Paint costs £7.50 a can and each can covers 5m2. How much will it cost Dasha to paint the wall?
👈🏼Masha put on weight this month. She weighed 75kg at the start of the month and 78kg at the end of the month. What was the percentage increase in her weight?
🖐🏼Simplify $\sqrt{8} \times \sqrt{2}$
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2023-03-23 21:37:58
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https://lqp2.org/node/804
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# The renormalized locally covariant Dirac field
Jochen Zahn
October 15, 2012
The definition of the locally covariant Dirac field is adapted such that it may be charged under a gauge group and in the presence of generic gauge and Yukawa background fields. We construct renormalized Wick powers and time-ordered products. It is shown that the Wick powers may be defined such that the current and the stress-energy tensor are conserved, and the remaining ambiguity is characterized. We sketch a variant of the background field method that can be used to determine the renormalization group flow at the one loop level from the nontrivial scaling of Wick powers.
open access link Rev. Math. Phys. 26 (2014) 1330012
@article{Zahn:2012dz, author = "Zahn, Jochen", title = "{The renormalized locally covariant Dirac field}", journal = "Rev. Math. Phys.", volume = "26", year = "2014", number = "1", pages = "1330012", doi = "10.1142/S0129055X13300124", eprint = "1210.4031", archivePrefix = "arXiv", primaryClass = "math-ph", reportNumber = "UWTHPH-2012-32", SLACcitation = "%%CITATION = ARXIV:1210.4031;%%" }
Keywords:
QFT on curved spacetimes, Dirac fields
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2019-10-16 01:52:27
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http://sdnjohnson.com/post/test-rmd/test-rmd/
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# Testing Rmd Support using a Schafer model
This is version 3 of my personal website. I’ve revised to a Hugo powered blogdown generated website for two reasons. First, it’s a static site generator, similar to Jekyll, serving to keep things simple and fast on the server side. Second, I’ve wanted to start blogging more, and I think the Rmarkdown integration with blogdown and Hugo will help smooth this out.
This post is a test post for using Rmarkdown to build a simple population dynamics model with process error, and generate some data with observation error. Pushing this post to github pages will require working out how TravisCI works, which I learned from David Selby’s Tea and Stats blog - a handy resource.
So, let’s get started by defining a function for the population dynamics model. I’m going to use a Schaefer surplus production model (Schaefer 1957). I like to define my Schafer models using optimal equilibrium values, known as $$B_{MSY}$$ and $$U_{MSY}$$ in the fisheries literature, as the leading parameters.
set.seed(123)
# Create a function to quickly call for generating biomass
prodMod <- function( Umsy = 0.06,
Bmsy = 42,
procSigma = 0.1,
obsSigma = 0.2,
nT = 100 )
{
# Vectors to hold biomass and catch
Bt <- numeric( length = nT )
Ct <- numeric( length = nT )
# Create a sequence of harvest rates
Utmult <- c(seq(0.1,2,length = 20),seq(2,1, length = 20), rep(1,nT - 40) )
Ut <- Utmult * Umsy
# And draw process errors
epst <- rnorm(nT-1, sd = procSigma)
# Initialise at unfished
Bt[1] <- 2 * Bmsy
# Take first year's catch
Ct[1] <- Ut[1] * Bt[1]
# Now loop and fill remaining years
for( t in 2:nT )
{
# Generate non-stochastic biomass
Bt[t] <- Bt[t-1] + 2 * Umsy * Bt[t-1] * (1 - Bt[t-1] / Bmsy / 2) - Ct[t-1]
Bt[t] <- Bt[t] * exp( epst[t-1])
# Take catch
Ct[t] <- Bt[t] * Ut[t]
}
# Draw observation errors
deltat <- rnorm(nT, sd = obsSigma)
# Generate absolute index of biomass
It <- Bt * exp(deltat)
# Return biomass, catch and model time dimension
outList <- list( Bt = Bt,
It = It,
Ct = Ct,
nT = nT )
}
# Generate the biomass
popSP <- prodMod()
# Plot the biomass and catch
plot( x = 1:popSP$nT, y = popSP$Bt, xlab = "Year", ylab = "Biomass and Catch",
ylim = c(0,max(popSP$Bt)), type = "l", col = "red", lwd = 2, las = 1 ) rect( xleft = 1:popSP$nT-.3, xright = 1:popSP$nT + .3, ybottom = 0, ytop = popSP$Ct, col = "grey40", border = NA )
points( x = 1:popSP$nT, y = popSP$It, pch = 21, col = "grey60",
cex = .8 )
Our figure shows a 2-way trip, which describes how the biomass declines at first (the first way), then comes back up as catch is decreased (the second way). This is an ideal data set for fisheries stock assessment, as there is contrast in the catch and biomass, making it informative about both the scale of the biomass, helping to estimate $$B_{MSY}$$, and the productivity of the stock, helping to estimate $$U_{MSY}$$.
The next part of this post will fit a TMB model to the data generated from the above biomass. I ran this same tutorial in a seminar for our lab group at SFU this past fall, so I’ll be copying and pasting the code in from there.
Schaefer, Milner B. 1957. “Some Considerations of Population Dynamics and Economics in Relation to the Management of the Commercial Marine Fisheries.” Journal of the Fisheries Board of Canada 14 (5). NRC Research Press: 669–81.
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2021-12-01 12:19:12
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https://math.meta.stackexchange.com/questions/16621/how-to-mention-someone-in-comments-if-his-her-name-doesnt-appear-after-typing-i
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# How to mention someone in comments if his\her name doesn't appear after typing its first letters?
Suppose there is a question of a member $A$. I want to mention this member $A$(so it appears in the comment as "@A"). sometimes,when I type the first letters of name "$A$", The name appears in a rectangle above and I click on the name so the member is mentioned. Many times, When I type the first letters , no thing happens!
What to do in this case? Can I mention him\her manually? How to do this? Suppose that his\her name was "$A\text{ } B$"(there is a space, like in "Maths Lover"). Should I type "@MathsLover" or "@Maths Lover"
Added: When I try to mention the member manually, his name disappear when I post my comment. How to deal with that?
• About your addition: meta.math.stackexchange.com/questions/6281/… Aug 21, 2014 at 13:40
• Spaces are removed from the display names for matching purposes. So to match Peter Smith you may use @pet, @peter, @peters, or @petersmith. Quote from here. Aug 21, 2014 at 13:44
When you comment under a post, the post author is always notified, so there is no need to alert her/him with an @-ping. If so far nobody except you and the post author have participated in the comment thread, an @author is automatically removed from the comment, since it is deemed clear whom you address with your comment. If you nevertheless want to explicitly address the post author, do that without an @. If other people have posted comments on the post, an @author is still not necessary to notify the author, but will not be removed, since it is now not a priori clear to whom your comment is addressed.
• Yes, the post author is always notified of comments on their posts, whether they contain an @-ping or not, and whether they are addressed at somebody else or not. (Unless the notification system is broken, which occasionally happens for some time.) Aug 21, 2014 at 13:10
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2022-07-02 14:04:07
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http://www.dxhx.pku.edu.cn/CN/abstract/abstract34210.shtml
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关于弯曲界面气液两相相平衡化学势相等热力学判据的讨论
• 发布日期:2016-12-20
• 通讯作者: 肖赛君 E-mail:jxddroc@126.com
• 基金资助:
国家自然科学基金(51404001);安徽省留学回国人员创新创业扶持计划资助项目[2016]
Discussion on Thermodynamic Criterion of Equal Chemical Potential for Phase Equilibrium between Gas and Liquid in Curved Interface
Sai-Jun XIAO*(),Jian LIU,Zhen-Xing YIN,Jun ZHANG
• Published:2016-12-20
• Contact: Sai-Jun XIAO E-mail:jxddroc@126.com
• Supported by:
国家自然科学基金(51404001);安徽省留学回国人员创新创业扶持计划资助项目[2016]
Abstract:
Thermodynamic criterion of equal chemical potential for phase equilibrium between gas and liquid in curved interface has been introduced in many textbooks and papers. In this paper, the process of the derivation of the criterion is proved to be wrong with analysis. In order to solve the problem, a proof procedure for the thermodynamic criterion of equal chemical potential is provided based on the definition of chemical potential. At the same time, a new thermodynamic criterion of two-phase equilibrium in curved interface is established using Gibbs interface thermodynamics and a new method for derivation of Kelvin equation is put forward based on the thermodynamic criterion of two-phase equilibrium. The new criterion which is derived directly from the second law of thermodynamics has a specific thermodynamic significance and clear physical model.
MSC2000:
• G64
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2022-01-17 07:23:53
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https://math.stackexchange.com/questions/541878/proving-that-gx-x1-xfx-is-uniformly-continuous-on-0-1-if-fx-is-c
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# Proving that $g(x)=x(1-x)f(x)$ is uniformly continuous on $(0,1)$ if $f(x)$ is continuous and bounded on $(0,1)$
Suppose that we have a function $f(x)$ that is continuous and bounded on $(0,1)$. Let us then define a function $g(x)=x(1-x)f(x)$. Prove that g(x) is continuous. I solved it through $\delta,\varepsilon$ definition of uniform continuity. However, I didn't need to use the fact that $f(x)$ was continuous; I basically implied through my proof that for $g(x)$ to be uniformly continuous, $f(x)$ only needed to be bounded. Which I think is wrong. Here is what I tried $$\vert g(x)-g(x_0)\vert$$$$\Rightarrow\vert x(1-x)f(x)-x_0(1-x_0)f(x_0)\vert$$$$\Rightarrow\vert x(1-x)f(x)-x(1-x)f(x_0)+x(1-x)f(x_0)-x_0(1-x_0)f(x_0)\vert$$$$\leq\vert x(1-x)\vert\vert f(x)-f(x_0)\vert+\vert f(x_0)\vert\vert x(1-x)-x_0(1-x_0)\vert$$
Since $f(x)$ is bounded we have that $\vert f(x)\vert\leq M$ for all x with M being some number. Thus we have that $$\vert x(1-x)\vert\vert f(x)-f(x_0)\vert+\vert f(x_0)\vert\vert x(1-x)-x_0(1-x_0)\vert$$$$\leq2M\vert x(1-x)\vert+M\vert x(1-x)-x_0(1-x_0)\vert$$ Afterwards, I used calculus to find the maximum of $x(1-x)$. I took derivative and set it equal to 0. This yielded me that $x(1-x)$ maximum is at $x=1/2$. Hence, $$2M\vert x(1-x)\vert+M\vert x(1-x)-x_0(1-x_0)\vert$$$$<2M\vert1/4\vert+M\vert1/4\vert=3M/4$$ Hence $\delta=4\varepsilon/3M$. Since $\delta$ does not depend on $x_0$, $g(x)$ is uniformly continuous. What did I do wrong? What are other methods to prove this?
• I do not see how you go from $3M/4$ to $\delta < 4\epsilon/3M$.
– user99914
Oct 27, 2013 at 19:57
• It should be an equal sign, sorry. If the question is more general: I thought it was similar as to how $\epsilon$ proofs for limits of sequences. Just that in this case, we are trying to express $\delta$. We were given a function, and found an expression, $\delta$, as to how far apart $x$ and $x_0$ are. We later related this to how far apart the $f(x)$ and $f(x_0)$ would be given delta. We want that $f(x)$ and $f(x_0)$ be within $\epsilon$. I am using the method presented in Ross and Lebl. Oct 27, 2013 at 20:23
• How do you show that $|g(x) - g(x_0)|\leq \epsilon$ for your choice of $\delta$?
– user99914
Oct 27, 2013 at 20:34
• We would expand |g(x)−g(x0)|≤ϵ and express the $\vert x-x_0\vert$ as $\delta$. Plug in the expression for $\delta$ and see whether |g(x)−g(x0)|≤ϵ holds. I am not quite sure what you're trying to get at. Could you elaborate? Oct 27, 2013 at 20:55
• Dont you come up with $|g(x) - g(x_0)| < 3M/4$? How do you come up with $\epsilon$?
– user99914
Oct 27, 2013 at 22:16
Hint: Let $\varepsilon > 0$ be given. Assume that $|f(x)| \le M$. Since $f$ is uniformly continuous on $\left[\dfrac{\varepsilon}{4M}, 1-\dfrac{\varepsilon}{4M}\right]$ you can choose $\tilde\delta$ such that $|f(x)-f(y)| < \varepsilon$ when $x,y \in \left[\dfrac{\varepsilon}{4M}, 1-\dfrac{\varepsilon}{4M}\right]$ and $|x-y| < \tilde\delta$. Now put $\delta := \min\left\{\tilde\delta, \dfrac{\varepsilon}{4M}\right\}$.
• How did you determine the interval? I understand that a continuous function is uniformly continuous on a closed interval but I am not quite sure why you chose $\varepsilon/4M$ Oct 27, 2013 at 21:05
• If $x<\dfrac{\varepsilon}{4M}$ and $|x-y|<\dfrac{\varepsilon}{4M}$, then $y<\dfrac{\varepsilon}{2M}$ and hence $|x(1-x)f(x)| \le \dfrac{\varepsilon}{2}$, $|y(1-y)f(y)| \le \dfrac{\varepsilon}{2}$. Similarly consider other cases. Oct 27, 2013 at 21:13
• I am confused, aren't we saying that x and y are elements of the closed set? And wouldn't that imply that both of them cannot be smaller than $\varepsilon/4M$? This is assuming that $\varepsilon/4M$ is smaller than 1, of course. But I cannot see how it would work even if $\varepsilon/4M$ is greater than 1. Oct 27, 2013 at 21:24
• For given $\varepsilon$ we chose some $\delta$. Now we have to prove that for any $x,y \in (0,1)$ with $|x-y|< \delta$ we get $|f(x)-f(y)| < \varepsilon$. If both of these points $\in \left[\dfrac{\varepsilon}{4M}, 1-\dfrac{\varepsilon}{4M}\right]$, then there is nothing to check. Otherwise, at least one of $x,y$ is outside that closed interval. Then try to consider all possible cases using my hints. Oct 27, 2013 at 21:31
Hint: If $f(x)$ is continuous in $(a,b)$ and there are $lim_{x\to a^+}f(x), lim_{x\to b^-}f(x)$, then $f(x)$ is uniformly continuous.
Since $x(1-x)$ and $f(x)$ are continuous in $(0,1)$, then $g(x)$ is continuous in $(0,1)$. Also, since there are $lim_{x\to 0^+}g(x)$ and $lim_{x\to 1^-}g(x)$, so $g(x)$ is uniformly continuous.
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2022-05-23 15:03:09
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https://mathoverflow.net/questions/328245/conceptual-explanation-for-chi-mathcala-g-chi-mathcal-m-1-times
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# Conceptual explanation for $\chi(\mathcal{A_g})=\chi(\mathcal M_{1} \times … \times \mathcal M_{g})$?
Let $$\mathcal{A_g}$$ be the moduli space of principally polarized abelian varieties over $$\mathbb C$$, $$\mathcal{M_g}$$ be the moduli space of smooth projective curves of genus $$g$$ over $$\mathbb C$$, and we regard both as Deligne-Mumford stacks or orbifolds. By computations we know their Euler characteristics :
$$\chi(\mathcal{A_g})=\chi(\mathcal M_{1})\chi(\mathcal M_{2})...\chi(\mathcal M_{g})=\prod_{k=1}^g \zeta(1-2k)$$.
Is there a conceptual explanation of this using Torelli map? Can we update it to an identity of Poincare polynomials?
• These Euler characteristics are not the alternating sums of Betti numbers and therefore aren't related to Poincare polynomials. For instance, they aren't integers. – Will Sawin Apr 17 at 2:12
• @WillSawin There are Poincaré polynomials associated with orbifolds. – sawdada Apr 17 at 2:53
• @WillSawin More precisely, one can replace $(-1)$ in $\chi(M/G) = \frac1{|G|}\sum_i\sum_{g\in G} (-1)^i\mathrm{Tr}_{H^i(M)}(g^*)$ by $t$ as in mathoverflow.net/questions/51993/… – sawdada Apr 17 at 14:25
• But that formula is for the Euler characteristic as an alternating sum of Betti numbers (see step 2 of Johannes Ebert's answer), not the other definition of the Euler characteristic for orbifolds, which gives $\chi(M/G) =\chi(M)/|G|$ and which is used in the zeta function identities you state. – Will Sawin Apr 17 at 14:50
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2019-07-19 06:56:26
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http://lists.llvm.org/pipermail/cfe-dev/2018-April/057674.html
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# [cfe-dev] Clang assembler missed features
Roberto E. Vargas Caballero via cfe-dev cfe-dev at lists.llvm.org
Mon Apr 16 07:41:09 PDT 2018
Hi,
I work in arm in an open source project called TF [1], and we support
compilation with clang. We want to get rid of GNU tools when a clang
toolchain is used, but we found some problems when we try to use the
clang assembler:
- .func, .endfunc and .struct directives are not supported
- substraction of relative labels doesn't produce absolute
values
.if (. - \begin) > 32)
generates 'error: expected absolute expression' message
Is clang going to support these features?
Best regards,
[1] https://github.com/ARM-software/arm-trusted-firmware
IMPORTANT NOTICE: The contents of this email and any attachments are confidential and may also be privileged. If you are not the intended recipient, please notify the sender immediately and do not disclose the contents to any other person, use it for any purpose, or store or copy the information in any medium. Thank you.
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2019-09-23 01:04:54
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https://phys.libretexts.org/TextMaps/Astronomy_and_Cosmology_TextMaps/Map%3A_Astronomy_(Impey)/12%3A_Properties_of_Stars/12.24__Stellar_Structure
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Skip to main content
$$\require{cancel}$$
# 12.24 Stellar Structure
There is an enormous difference between a puny red dwarf on the main sequence, barely hot enough to sustain fusion reactions, and a massive blue main sequence star, pumping out energy at a billions of times faster rate. In this spectrum, the Sun occupies the middle ground. Can we understand these differences in terms of simple physics? Yes. The structure of a main sequence star of any mass is governed by four considerations: conservation of mass, conservation of energy, hydrostatic equilibrium or the balance between pressure and gravity, and the way energy is transported or the relation between energy flow and the temperature gradient. Notice that rate of energy generation by fusion is not a primary consideration! That’s what let theorists like Eddington and Russell work out the structure of stars in the 1930s, before the fusion process was fully understood.
We’ve seen that main sequence stars have a luminosity that increases strongly with increasing mass, leading to that huge difference between high and low mass stars on the main sequence. For high mass stars, the primary way energy travels out through the star is as radiation. For low mass stars, it travels outwards by convection, or wholesale motions of the plasma.The interiors of stars are extremely hot, millions of degrees. The fall-off to surface temperatures thousands or tens of thousands of degrees takes place in a thin outer shell. Energy generation by fusion takes place in a small central region, involving a small fraction of the total mass of the star. The rapid fall-off of nuclear energy generation reflects the fact that it is very sensitive to temperature. Stars less massive than the Sun generate most of the energy by the proton-proton chain. Stars most massive than the Sun generate most of their energy by the CNO Cycle. When the details of energy transport are calculated, it turns out that massive stars are hotter in cores but actually less dense than low mass stars.
Stars are in equilibrium while they are on the main sequence. Suppose the fusion rate increases slightly. The temperature and pressure would increase, the core would expand slightly, which would lower the density and temperature, decreasing the fusion rate. so a feedback mechanism stops the fusion rate from skyrocketing. The reverse argument works as well. Stars are steady sources of radiation and light. But suppose there was no energy generation in the core. In this case, the pressure would be high and the core would be hotter than the outer envelope. Energy would escape by radiation or convection and so the core would shrink a bit by gravity. That would make it hotter, so even more energy would escape. Now it's a positive feedback loop so the effect is amplified not suppressed. The core collapses! That’s why dramatic changes occur when a main sequence star exhausts its hyrogen, and any time a star reaches the end of its nuclear fuel.
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2017-11-18 15:55:07
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https://tex.stackexchange.com/questions/572478/how-to-define-a-unicode-math-symbol-available-in-a-unicode-math-font-in-terms-of
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# How to define a unicode math symbol available in a unicode math font in terms of a hexadecimal value?
Given that I have a unicode math font with a glyph that I want to use in math mode, how can I, using the unicode-math package, define the glyph? I can get this to kind of work using \DeclareUTFcharacter, but with a resulting "invalid in math mode" warning.
\documentclass{article}
\usepackage{xunicode}
\usepackage{unicode-math}
\setmathfont{XITSMath-Regular.otf}
\DeclareUTFcharacter[\UTFencname]{x24C7}{\circledR} % "LaTeX Warning: Command \circledR invalid in math mode"
%\UnicodeMathSymbol{"024C7}{\circledR} {\mathrel}{circled R}
%\__um_process_symbol_noparse:nnn {"024C7}{\circledR} {\mathbin}
\begin{document}
circleddash: $\circleddash$
\\circledparallel: $\circledparallel$
\\circledR: $\circledR$
\end{document}
I'm not sure there's an official interface; anyway, this works.
\documentclass{article}
\usepackage{unicode-math}
\setmathfont{XITSMath-Regular.otf}
\Umathchardef\circledR="0 \symsymbols "24C7
\begin{document}
circleddash: $\circleddash$
circledparallel: $\circledparallel$
circledR: $\circledR$
\end{document}
• Thank you @egreg—that's very helpful. Yes, an interface would be useful (as you suggested some 5 years ago)... if someone had the time to implement one. – Daniel J. Greenhoe Nov 26 '20 at 15:43
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2021-04-17 08:38:51
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https://eprint.iacr.org/2019/069
|
### Quantum Indistinguishability of Random Sponges
Jan Czajkowski, Andreas Hülsing, and Christian Schaffner
##### Abstract
In this work we show that the sponge construction can be used to construct quantum-secure pseudorandom functions. As our main result we prove that random sponges are quantum indistinguishable from random functions. In this setting the adversary is given superposition access to the input-output behavior of the construction but not to the internal function. Our proofs hold under the assumption that the internal function is a random function or permutation. We then use this result to obtain a quantum-security version of a result by Andreeva, Daemen, Mennink, and Van Assche (FSE'15) which shows that a sponge that uses a secure PRP or PRF as internal function is a secure PRF. This result also proves that the recent attacks against CBC-MAC in the quantum-access model by Kaplan, Leurent, Leverrier, and Naya-Plasencia (Crypto'16) and Santoli, and Schaffner (QIC'16) can be prevented by introducing a state with a non-trivial inner part. The proof of our main result is derived by analyzing the joint distribution of any $q$ input-output pairs. Our method analyzes the statistical behavior of the considered construction in great detail. The used techniques might prove useful in future analysis of different cryptographic primitives considering quantum adversaries. Using Zhandry's PRF/PRP switching lemma we then obtain that quantum indistinguishability also holds if the internal block function is a random permutation.
Available format(s)
Category
Secret-key cryptography
Publication info
Preprint. Minor revision.
Keywords
Symmetric cryptographykeyed spongesindistinguishabilityquantum securitymessage-authentication codes
Contact author(s)
j czajkowski @ uva nl
andreas @ huelsing net
c schaffner @ uva nl
History
Short URL
https://ia.cr/2019/069
CC BY
BibTeX
@misc{cryptoeprint:2019/069,
author = {Jan Czajkowski and Andreas Hülsing and Christian Schaffner},
title = {Quantum Indistinguishability of Random Sponges},
howpublished = {Cryptology ePrint Archive, Paper 2019/069},
year = {2019},
note = {\url{https://eprint.iacr.org/2019/069}},
url = {https://eprint.iacr.org/2019/069}
}
Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.
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2022-05-26 03:02:10
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https://electronics.stackexchange.com/questions/432551/how-to-make-momentary-connection-on-power-on
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# How to make momentary connection on power on?
I have a mini-computer Beelink AP34 with momentary power button and I need to simulate short pressing of this button once it is connected to the power (12V DC).
The reason is that it doesn't start automatically after power loss and does not support Wake-on-LAN or anything - I always have to physically press that button after connecting power adapter.
• Is the "mini-computer" powered by this 12 V supply? Or something else? – jonk Apr 14 at 20:16
• check the bios settings for restart on powerfail or some similar setting – jsotola Apr 14 at 20:23
• @jonk - yes it is powered by AC-DC 12V adapter. – verglor Apr 14 at 20:26
• @jsotola unfortunately BIOS does not have this setting - I believe hardware mod is inevitable. – verglor Apr 14 at 20:27
• So, all you need is a delayed power-on signal (which could start with a slow RC charging process) followed by something with hysteresis to "snap-on" a switch, so to speak. You could probably use a relay, safely. But if you want to avoid a relay then what do you know about the button? What exactly is it attached to? (You need to work out precisely the circuit connections the switch is tied to or else you may be forced to use a relay to be safer.) A latching relay would permit lower continual power consumption. But a regular relay could also be used if the power is acceptable. – jonk Apr 14 at 20:37
You wrote that, "I am total rookie in electronics, so I need it to be as simple as possible." Since none of my betters (I'm just a hobbyist without even so much as one day's class of DC electronics training) has yet bothered to provide something I'll try to follow your guiding words and provide something simple and easy and relatively available.
simulate this circuit – Schematic created using CircuitLab
Once the 555 becomes active, both the $$\\text{THRESHOLD}\$$ and $$\\text{TRIGGER}\$$ pins should be low enough (because $$\C_1\$$ is holding them both low) to start that the output will be active-HI and the relay will be engaged. I've specified one particular $$\12\:\text{V}\$$ relay you might consider using. It's a TE Connectivity IMB06CTS and is a signal relay (instead of a power relay) with a coil resistance of about $$\1\:\text{k}\Omega\$$. The 555 can easily drive it, directly.
Eventually, $$\C_1\$$ will charge up sufficiently that those two pins will be above the needed threshold and the output will go active-LO and the relay will be disabled.
I've set up $$\R_1\$$ and $$\C_1\$$ to provide about $$\750\:\text{mS}\$$ duration for the relay. That's a "long-press" of your button, so feel free to shorten that up a bit by reducing the values of either the resistor or capacitor. For example, you might use $$\R_1=270\:\text{k}\Omega\$$, instead. Or $$\C_1=220\:\text{nF}\$$, instead. Either change will probably work fine. So that gives you an idea of the range of change you might consider.
Just in case it's not entirely clear to you, the switch within the relay (shown in the diagram) should be used in parallel with your existing manual momentary switch. No need to remove the manual switch, if you want to keep it and use it. This relay switch simply bypasses it to allow for an automatic restart, as well.
• Thank You very much. This seems to be exactly what I was looking for. I am just interested if it is possible to replace IMB06CTS with any other 12V signal realy - preferably available on aliexpress or any other eshop with cheap intl shipping. Maybe something like this Omron G5V-2 ? – verglor Apr 16 at 21:32
• @verglor The 555 output can probably handle almost any small relay. Since all you are doing is emulating a switch, they don't need to have large contacts. The Omron you mentioned has about $\frac14$ the resistance and will require four times the current but the 555 can handle that, too, just fine. So I think you should feel okay with it. – jonk Apr 16 at 21:41
• Hi @jonk, I have hooked it up on breadboard and it's working prefectly. However I have some silly questions: 1) what is the purpose of C2 ? (I didn't have capacitor with so low capacity so I omitted it and it still works) 2) how can I introduce a little delay before the relay "pushes the button" ? – verglor Apr 24 at 13:08
• @verglor Add a big capacitor between ground and the reset pin. Very big, like maybe $47\:\mu\text{F}$. It takes a lot because that pin can maximally source as much as $1.5\:\text{mA}$. Adjust up or down from there, once you test it out, to get the desired delay. – jonk Apr 24 at 16:05
• @verglor $C_2$ is just a standard, recommended value to help stabilize a resistor divider inside. You could probably eliminate it if it bothers you. – jonk Apr 24 at 16:08
The 555 is a great idea, but it's worth pointing out that an arduino nano could also do the job. The power switch likely jumps a pin pulled to 5v to ground. In such case, a 5v powered arduino could do the job. It may seem like overkill, but it's a software driven solution that requires no soldering - just an arduino nano with pins soldered (around five bucks on ebay) and three conductors (for gnd, 5v, and pin) of a 40 conductor preassembled jumper. Program the arduino to pull a pin low momentarily two seconds after reset, connect that pin to the pwr_on pin on the motherboard, and there ya go. You can likely find a manual on the net that will give you the required header pins on the motherboard to provide 5v and gnd and pwr_on. This means no soldering, maybe not even needing a vom to measure things out.
Edit: apparently they have a bios revision to solve this problem. I'd suggest going to their support site and doing a bios update.
• I have found in the forum that there should be BIOS update. But it is not generally available and when I asked support to provide it, they said, that they don't have BIOS with auto start but that they support WOL. It's strange - however my current version of BIOS does not support either feature. – verglor Apr 16 at 21:22
• However your suggestion of arduino solution is interesting indeed. Thank You. I'll give it a try if I will not be able to get BIOS update. – verglor Apr 16 at 21:36
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2019-05-26 17:43:23
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https://hsm.stackexchange.com/questions/11259/who-first-gave-a-definition-of-congruent-triangles
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# Who first gave a definition of congruent triangles?
Who was the first to define congruent triangles? I couldn't find the definition in Euclid's Elements.
• It’s in Common Notion 4 in Book I of Euclid. – Michael E2 Dec 25 '19 at 6:38
• @MichaelE2 "Things which coincide with one another equal one another" is not exactly a definition of congruent triangles. The closest he has to a definition is written into Proposition 4 of the same book:"If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, then they also have the base equal to the base, the triangle equals the triangle, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides." – Conifold Dec 25 '19 at 6:47
• I would have added some more but I’m on a phone. it messed up my message. The phrase is epharmozonta ep’ allela that corresponds to congruent – Michael E2 Dec 25 '19 at 6:51
It depends on how the question is interpreted. Common notion 4, "things which coincide with one another equal one another", is not exactly a definition of congruent triangles or even of congruence generally. Likely due to Platonist strictures, Euclid deliberately avoids using congruence in his demonstrations. The idea of moving and superimposing figures Plato judged as "corrupting the good of geometry", which is why we do not find "workman tools" like straightedge and compass, or mechanically generated curves, in the Elements either, see When were the concepts of pure and applied Mathematics introduced?
Many Euclidean demonstrations display proliferation of auxiliary triangles where the use of congruence would have simplified the clutter. But at the very beginning Euclid has no choice, as he needs tests of equality for his auxiliary triangles. The first such test is Proposition I.4:
"If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, then they also have the base equal to the base, the triangle equals the triangle, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides."
All sides and angles are equal to each other is a way to define "congruence of triangles" without congruence. Euclid could have just made I.4 another postulate, but as he wanted to give a demonstration he had to resort to superimposing. Even so, the use of imposition is buried inside the demonstration and unacknowledged. Joyce comments on this in a note to I.26 of his translation of the Elements:
"Euclid’s congruence theorems are I.4 (side-angle-side), I.8 (side-side-side), and this one, I.26 (side and two angles). Calling them congruence theorems is anachronistic, since Euclid did not explicitly use the concept of congruence. We would say that two triangles ABC and DEF are congruent if the angles A, B, and C equal the angles D, E, and F respectively, and the sides AB, BC, and AC equal the sides DE, EF, and DF respectively, and the triangle ABC equals the triangle DEF (by which is meant that they have the same area).
The word "congruent" for superimposing (superpono) appears in 16-17th century Latin translations of Euclid, whose authors had no preconceptions about motion. "Congruent" replaces "coincide" in Common notion 4. Clavius in Euclidis Elementorum (1589) writes, for example:
"Hinc enim fit, ut aequalitas angulorum ejusdem generis requirat eandem inclinationem linearum, ita ut lineae unius conveniant omnino lineis alterius, si unus alteri superponatur. Ea enim aequalia sunt, quae sibi mutuo congruunt."
"For want of this, equality of the angles of the same kind requires the same inclination of the lines, and lines to lines coincide when placed on top of each other. For those things are equal, that are congruent with each other." [translation mine]
According to Miller's Earliest Known Uses the word is first used systematically in something like the modern sense only by Leibniz in 1679:
"As a more technical term for a relation between figures, congruent seems to have originated with Gottfried Wilhelm Leibniz (1646-1716), writing in Latin and French. His manuscript "Characteristica Geometrica" of August 10, 1679, is in his Gesammelte Werke, dritte Folge: mathematische Schriften, Band 5. On p. 150 he says that if a figure can be applied exactly to another without distortion, they are said to be congruent...
His Figure 39 shows two radii of a circle, with the center labelled both A and C. Later (p. 154) he points out that "congruent" is the same as "similar and equal." He used "congruent" in the modern (Hilbert) sense, applied to line segments and various other things as well as triangles.
Shortly afterwards, on September 8, 1679, he included a similar definition in a letter to Hugens (sic) van Zulichem. In his ges. Werke etc. as above, volume 2, p. 22, he illustrates congruence with a pair of triangles, and says that they "peuvent occuper exactement la meme place, et qu'on peut appliquer ou mettre l'un sur l'autre sans rien changer dans ces deux figures que la place." [Ken Pledger and Julio González Cabillón]".
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2021-04-14 10:59:20
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https://tex.stackexchange.com/questions/467006/set-coordinate-boundaries-of-tikz-picture
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# Set coordinate boundaries of tikz picture
From my small experience with tikz in Latex, it seems that the boundary of a tikz image is set automatically based on its contents. This caused me some trouble when I was trying to scale and align multiple tikz images.
Is there a way to set the boundary manually? For instance perhaps I want an image with nodes at (1,1) and (2,1) and a boundary which is rectangular with corners at (0,0) and (3,2). How would this be achieved?
• give a complete example! – user2478 Dec 22 '18 at 17:30
• \clip (-0,05) rectangle (3,2); welcome to tex.se! – Zarko Dec 22 '18 at 17:30
• \path[use as bounding box] (0,0) rectangle (1,1); See p. 175 of the pgfmanual for more details. – user121799 Dec 22 '18 at 17:46
• Thanks all! Zarko's answer did the trick. @Zarko: Did you want to move your comment to an answer? – Erewhon Dec 22 '18 at 17:51
• of course i can, but this can be easy to do if you provide some mwe ... :-) – Zarko Dec 22 '18 at 17:52
use of clip you can see from the following example from TikZ & PGF Manuall, v 3.0.1A, page 140:
\begin{tikzpicture}
\clip (-2,-2) rectangle (2,2);
\draw [name path=curve 1] (-2,-1) .. controls (8,-1) and (-8,1) .. (2,1);
\draw [name path=curve 2] (-1,-2) .. controls (-1,8) and (1,-8) .. (1,2);
\fill [name intersections={
of=curve 1 and curve 2,
by={[label=center:a],[label=center:...],[label=center:i]}}];
\end{tikzpicture}
description you can find in chapter 15 Actions on Paths, page 164 of aforementioned manual.
It is possible to use use as bounding box (see page 175 of the 3.0.1a manual) and its shortcut \useasboundingbox (see page 164)
Example taken from page 236 of the manual:
\documentclass[tikz]{standalone}
\begin{document}
\begin{tikzpicture}[auto]
\draw[help lines,use as bounding box] (0,-.5) grid (4,5);
\draw (0.5,0) .. controls (9,6) and (-5,6) .. (3.5,0)
node foreach \pos in {0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1}
[pos=\pos,swap,fill=red!20] {\pos}
node foreach \pos in {0.025,0.2,0.4,0.6,0.8,0.975}
[pos=\pos,fill=blue!20] {\pos};
\end{tikzpicture}
\end{document}
Just include the desired margin coordinates in any path and the resulting tikzpicture will include it in final bounding box. In following picture, a phantom path is drawn and included in final box (shown in black with show background rectangle).
\documentclass[tikz,border=2mm]{standalone}
\usetikzlibrary{positioning, backgrounds}
\begin{document}
\begin{tikzpicture}[show background rectangle]
\draw[red] (1,1) rectangle ++(1,1);
\path(0,0) -- (3,2);
\end{tikzpicture}
\end{document}
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2021-07-25 20:10:17
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https://nbviewer.jupyter.org/url/schmidt.nuigalway.ie/cs4423/networks22.ipynb
|
### CS4423 - Networks¶
Prof. Götz Pfeiffer
School of Mathematics, Statistics and Applied Mathematics
NUI Galway
# Lecture 22: Configuration Model¶
In [ ]:
import numpy as np
import pandas as pd
import networkx as nx
import matplotlib.pyplot as plt
### The Configuration Model¶
In principle, random graph can be generated in such a way that they have a prescribed degree sequence.
Idea: Choose numbers $d_i$, $i \in X$, so that $\sum d_i = 2m$ is an even number.
Then regard each degree $d_i$ as $d_i$ stubs (half-edges) attached to node $i$.
Compute a random matching of pairs of stubs and build a graph on $X$ with those (full) edges.
Example. Suppose that $X = \{1, 2, 3, 4, 5\}$ and that we want those nodes to have degrees $d_1 = 3$, $d_2 = 2$ and $d_3 = d_4 = d_5 = 1$.
This gives a list of stubs $(1, 1, 1, 2, 2, 3, 4, 5)$ where each node $i$ appears as often as its degree $d_i$ requires.
A random shuffle of that list is $(1, 3, 4, 1, 2, 1, 5, 2)$.
One way to construct a matching is to simply cut this list in half and match entries of the first half with corresponsing entries in the second half. $$\begin{array}{cccc} 1 & 3 & 4 & 1\\ 2 & 1 & 5 & 2 \end{array}$$ Note that $\sum d_i = 8 = 2m$ yields $m = 4$ edges ...
### A Quick Implementation¶
In [ ]:
degrees = [3, 2, 1, 1, 1]
Recall that, in Python, list addresses start at $0$, and networkx default node names do likewise. Let's adopt this convention here.
Now entry $3$ in position $0$ of the list degrees stands for $3$ entries $0$ in the list of stubs, to be constructed. Entry $2$ in position $1$ stands for $2$ entries $1$ in the list of stubs and so on. In general, entry $d$ in position $i$ stands for $d$ entries $i$ in the list of stubs.
Python's list arithmetic (using m * a for m repetitions of a list a and a + b for the concatenation of lists a and b) can be used to quickly convert a degree sequence into a list of stubs as follows.
In [ ]:
stubs = [degrees[i] * [i] for i in range(len(degrees))]
stubs
In [ ]:
stubs = sum(stubs, [])
stubs
Let's call this process the stubs list of a list of integers and wrap it into a python function
In [ ]:
def stubs_list(a):
return sum([a[i] * [i] for i in range(len(a))], [])
In [ ]:
stubs_list(degrees)
How to randomly shuffle this list? The wikipedia page on random permutations recommends a simple algorithm for shuffling the elements of a list a in place.
In [ ]:
from random import randint
def knuth_shuffle(a):
l = len(a)-1
for k in range(l):
j = randint(k, l)
a[j], a[k] = a[k], a[j]
In [ ]:
a = [1,2,3]
knuth_shuffle(a)
a
Let's test whether this shuffle produces uniformly random outcomes ...
In [ ]:
shuffles = {}
for i in range(1000):
a = [1,2,3]
knuth_shuffle(a)
key = tuple(a)
shuffles[key] = shuffles.get(key, 0) + 1
print(shuffles)
But python's random module already contains a function shuffle which does exactly this.
In [ ]:
from random import shuffle
In [ ]:
a = [1,2,3]
shuffle(a)
a
Let's test whether this shuffle produces uniformly random outcomes ...
In [ ]:
shuffles = {}
for i in range(1000):
a = [1,2,3]
shuffle(a)
key = tuple(a)
shuffles[key] = shuffles.get(key, 0) + 1
print(shuffles)
So we shuffle the stubs ...
In [ ]:
shuffle(stubs)
stubs
Then we match pairs, by cutting the list of stubs into halves and transposing the resulting array of 2 rows ...
In [ ]:
m = len(stubs) // 2
In [ ]:
edges = [stubs[:m], stubs[m:]]
edges
In [ ]:
edges = list(zip(*edges))
edges
In [ ]:
G = nx.Graph(edges)
In [ ]:
G.number_of_edges()
In [ ]:
nx.draw(G, with_labels=True, node_color='y')
In [ ]:
G.edges()
All in all, a configuration model can be built as follows.
In [ ]:
def configuration(degrees):
m = sum(degrees) // 2 # should check if sum(degs) is even ...
stubs = stubs_list(degrees)
shuffle(stubs)
edges = list(zip(stubs[:m], stubs[m:]))
return nx.Graph(edges)
In [ ]:
G = configuration([3,2,1,1,1])
nx.draw(G, with_labels=True, node_color='y')
print(G.edges())
Now create a power degree distribution ... and generate a graph. Use $\gamma = 2$, since I know that $\zeta(2) = \pi^2/6$ ...
In [ ]:
from math import pi
gamma = 2
c = 6/pi/pi
p = [0] + [c * k**(-gamma) for k in range(1,35)]
print(p, sum(p))
Note how $p_0 = 0$ was explicitly prepended to the list p. Turn this now into a degree histogram on, say, $n = 50$ nodes.
In [ ]:
d = [int(50 * x) for x in p]
print(d)
print(sum(d))
Then recover the degree sequence from the histogram. Incidently, this works exactly as the transformation of the degree sequence above into a list of stubs (why?).
In [ ]:
print(stubs_list(d))
In [ ]:
G = configuration(stubs_list(d))
In [ ]:
G.edges()
In [ ]:
nx.draw(G, with_labels=True, node_color='y')
In [ ]:
nx.degree_histogram(G)
Try again ... This time with $\gamma= 2.5$, an explict value of $c = 2.5$, and $p[1]$ corrected so that $\sum p_k = 1$.
In [ ]:
gamma = 2.5
c = 2.5
p = [0] + [c * k**(-gamma) for k in range(1,35)]
p[1] = p[1] - (sum(p) - 1)
print(p)
print(sum(p))
Lift the degree numbers by $1/2$ so that they are rounded, rather than cut off when converted into integers. Also reduce number of nodes to $25$ for better drawings ...
In [ ]:
d = [int(25 * x + 0.5) for x in p]
print(d)
print(sum(d))
In [ ]:
print(stubs_list(d))
In [ ]:
G = configuration(stubs_list(d))
In [ ]:
G.edges()
In [ ]:
nx.draw(G, with_labels=True, node_color='y')
In [ ]:
nx.degree_histogram(G)
In networkx, configuration models can be generated with the function nx.configuration_model.
In [ ]:
G = nx.configuration_model(degrees)
nx.draw(G, with_labels=True, node_color='m')
In [ ]:
|
2019-12-09 01:46:27
|
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https://www.biostars.org/p/9488992/
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BCFtools Allelic Depth format nowhere explained?
1
0
Entering edit mode
11 weeks ago
chrisgr ▴ 20
It has this format e.g. AD=262,18,0 - What number shows the depth of the REF and what is the ALT? and what is the '0'? I found 1 form where someone said 262 is the REF and 18 is the ALT. Is that correct?
I would also like to know why this format is explained nowhere in the manuals.. It seems to me like something everyone would be interested in.
-Chris
Allelic Samtools BCFtools Depth AD • 349 views
0
Entering edit mode
I noticed that it follows the same format as the ref and alt collumn of the vcf so, 262 indecates the REF, the 18 the most occuring alt, and the 0 the depth of the 2nd most occuring alt
1
Entering edit mode
11 weeks ago
is a tri-allelic variant. 262 reads are REF, 18 are ALT n°1, 0 are ALT n°2
0
Entering edit mode
But no, not in the VCF header. It says only this:
Thank you for the explanation!
-Chris
0
Entering edit mode
FORMAT=<ID=AD,Number=R,Type=Integer,Description="Allelic depths"> . So all the information you need is here. See the VCF spec for the meaning of Number=R...
0
Entering edit mode
Oh I see!
Thank you!
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2021-11-29 15:03:05
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https://www.transtutors.com/questions/a-56-kg-bungee-jumper-jumps-off-a-bridge-and-undergoes-simple-harmonic-motion-if-the-1469334.htm
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a 56 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. if the period of osci 1 answer below »
a 56 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. if the period of oscillations is 11.2 s, what is the spring constant of the bungee cord?
First, we calculate the angular frequency $$\omega=\frac2\piT=\frac2\times3.1411.2=0.561 rad/s$$ Beside, $$\omega=\sqrt\frackm$$ k is the spring constant. m...
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2021-03-02 14:53:48
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https://scikit-learn.org/stable/modules/generated/sklearn.metrics.precision_recall_fscore_support.html
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# sklearn.metrics.precision_recall_fscore_support¶
sklearn.metrics.precision_recall_fscore_support(y_true, y_pred, beta=1.0, labels=None, pos_label=1, average=None, warn_for=(‘precision’, ’recall’, ’f-score’), sample_weight=None)[source]
Compute precision, recall, F-measure and support for each class
The precision is the ratio tp / (tp + fp) where tp is the number of true positives and fp the number of false positives. The precision is intuitively the ability of the classifier not to label as positive a sample that is negative.
The recall is the ratio tp / (tp + fn) where tp is the number of true positives and fn the number of false negatives. The recall is intuitively the ability of the classifier to find all the positive samples.
The F-beta score can be interpreted as a weighted harmonic mean of the precision and recall, where an F-beta score reaches its best value at 1 and worst score at 0.
The F-beta score weights recall more than precision by a factor of beta. beta == 1.0 means recall and precision are equally important.
The support is the number of occurrences of each class in y_true.
If pos_label is None and in binary classification, this function returns the average precision, recall and F-measure if average is one of 'micro', 'macro', 'weighted' or 'samples'.
Read more in the User Guide.
Parameters: y_true : 1d array-like, or label indicator array / sparse matrix Ground truth (correct) target values. y_pred : 1d array-like, or label indicator array / sparse matrix Estimated targets as returned by a classifier. beta : float, 1.0 by default The strength of recall versus precision in the F-score. labels : list, optional The set of labels to include when average != 'binary', and their order if average is None. Labels present in the data can be excluded, for example to calculate a multiclass average ignoring a majority negative class, while labels not present in the data will result in 0 components in a macro average. For multilabel targets, labels are column indices. By default, all labels in y_true and y_pred are used in sorted order. pos_label : str or int, 1 by default The class to report if average='binary' and the data is binary. If the data are multiclass or multilabel, this will be ignored; setting labels=[pos_label] and average != 'binary' will report scores for that label only. average : string, [None (default), ‘binary’, ‘micro’, ‘macro’, ‘samples’, ‘weighted’] If None, the scores for each class are returned. Otherwise, this determines the type of averaging performed on the data: 'binary': Only report results for the class specified by pos_label. This is applicable only if targets (y_{true,pred}) are binary. 'micro': Calculate metrics globally by counting the total true positives, false negatives and false positives. 'macro': Calculate metrics for each label, and find their unweighted mean. This does not take label imbalance into account. 'weighted': Calculate metrics for each label, and find their average weighted by support (the number of true instances for each label). This alters ‘macro’ to account for label imbalance; it can result in an F-score that is not between precision and recall. 'samples': Calculate metrics for each instance, and find their average (only meaningful for multilabel classification where this differs from accuracy_score). warn_for : tuple or set, for internal use This determines which warnings will be made in the case that this function is being used to return only one of its metrics. sample_weight : array-like of shape = [n_samples], optional Sample weights. precision : float (if average is not None) or array of float, shape = [n_unique_labels] recall : float (if average is not None) or array of float, , shape = [n_unique_labels] fbeta_score : float (if average is not None) or array of float, shape = [n_unique_labels] support : int (if average is not None) or array of int, shape = [n_unique_labels] The number of occurrences of each label in y_true.
References
Examples
>>> from sklearn.metrics import precision_recall_fscore_support
>>> y_true = np.array(['cat', 'dog', 'pig', 'cat', 'dog', 'pig'])
>>> y_pred = np.array(['cat', 'pig', 'dog', 'cat', 'cat', 'dog'])
>>> precision_recall_fscore_support(y_true, y_pred, average='macro')
...
(0.22..., 0.33..., 0.26..., None)
>>> precision_recall_fscore_support(y_true, y_pred, average='micro')
...
(0.33..., 0.33..., 0.33..., None)
>>> precision_recall_fscore_support(y_true, y_pred, average='weighted')
...
(0.22..., 0.33..., 0.26..., None)
It is possible to compute per-label precisions, recalls, F1-scores and supports instead of averaging:
>>> precision_recall_fscore_support(y_true, y_pred, average=None,
... labels=['pig', 'dog', 'cat'])
...
(array([0. , 0. , 0.66...]),
array([0., 0., 1.]), array([0. , 0. , 0.8]),
array([2, 2, 2]))
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2018-11-17 02:32:11
|
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https://dsp.stackexchange.com/tags/deconvolution/new
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We’re rewarding the question askers & reputations are being recalculated! Read more.
# Tag Info
How can I justify the expression of G′(ν)? That's easy enough. Denominator is the sum of the signal energy $|X(\omega)|^2$ and the noise energy $\lambda ^2$. If the signal energy is significantly larger, then the whole expression simplifies to $G(\omega)$. If the noise is larger, we can't do a anything useful with the information and the $1/ \lambda ^2$ ...
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2019-11-17 12:34:53
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https://www.scielo.br/j/rbeaa/a/8nwBcS6pHjrQmSRRYp8KCpD/?lang=en
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# HIGHLIGHTS:
The sapodilla leaf area can be estimated using a nondestructive method based on allometric equations.
Models that use the product of the leaf length and width present the best criteria for estimating the leaf area.
The equations LA = 0.664 × LW1.018 and LA = 0.713 × LW accurately and quickly estimate the sapodilla leaf area.
# ABSTRACT
The leaf area is a parameter of fundamental importance in studies on plant growth and physiology. The objective of this study was to build allometric equations for the accurate and fast estimation of sapodilla leaf areas. In total, 250 leaves of different shapes and sizes were collected from sapodilla matrices trees growing at the Universidade Federal Rural do Semi-Árido, Mossoró-RN, Brazil. For each leaf, the length, width, product of length and width (LW), product of length and length, product of width and width, and leaf area were measured. Linear and nonlinear models were used to construct the allometric equations. The best equations were chosen on the basis of the following criteria: the highest coefficient of determination, Pearson’s linear correlation coefficient, and Willmott’s index of agreement; and the lowest Akaike information criterion and root mean square error. It was verified that the models that used the LW value presented the best criteria for estimating the leaf area. Specifically, the equations ŷ = 0.664 × LW1.018 and ŷ = 0.713 × LW, which use LW values, are the most suitable for estimating the leaf area of sapodilla quickly and accurately.
Key words:
Manilkara zapota L.; biometrics; regression models; Sapotaceae
# RESUMO
Palavras-chave:
Manilkara zapota L.; biometria; modelos de regressão; Sapotaceae
# Introduction
Among the fruits of nutritional importance, sapodilla (Manilkara zapota L.; family Sapotaceae) is an exotic species with known antioxidative properties. The sapodilla trees are cultivated widely in pantropical regions for their wood (used for various purposes, including the creation of artisanal products), fresh fruits, and latex (Moura et al., 2019Moura, B. I. de V.; Araújo, B. P. L. de; Sa, R. D.; Randau, K. P. Pharmacobotanical study of Manilkara zapota (L.) P. Royen (Sapotaceae). Brazilian Journal of Pharmaceutical Sciences, v.55, p.1-10, 2019. https://doi.org/10.1590/s2175-97902019000117227
https://doi.org/10.1590/s2175-9790201900...
). For the continued expansion of sapodilla, studies are needed to estimate its growth, physiology, and productivity.
The leaf area (LA), which is an essential parameter for studies of plant physiology and ecology, is directly related to vital plant processes, such as photosynthetic efficiency, water balance, respiration, and the interception of light energy (Tondjo et al., 2015Tondjo, K.; Brancheriau, L.; Sabatier, S.; Kokutse, A. D.; Akossou, A.; Kokou, K.; Fourcaud, T. Non-destructive measurement of leaf area and dry biomass in Tectona grandis. Trees, v.29, p.1625-1631, 2015. https://dx.doi.org/10.1007/s00468-015-1227-y
https://dx.doi.org/10.1007/s00468-015-12...
). LA measurements can be obtained either directly (using a digital meter or scanner) or indirectly (Zhang, 2020Zhang, W. Digital image processing method for estimating leaf length and width tested using kiwifruit leaves (Actinidia chinensis Planch). PloS one, v.15, p.1-14, 2020. https://doi.org/10.1371/journal.pone.0235499
https://doi.org/10.1371/journal.pone.023...
). The direct methods involve the destruction of samples, which renders successive measurements unfeasible and incurs high costs (Liu et al., 2017Liu, Z.; Zhu, Y.; Li, F.; Jin, G. Non-destructively predicting leaf area, leaf mass and specific leaf area based on a linear mixed-effect model for broadleaf species. Ecological Indicators, v.78, p.340-350, 2017. https://doi.org/10.1016/j.ecolind.2017.03.025
https://doi.org/10.1016/j.ecolind.2017.0...
). By contrast, the indirect method involves the application of regression models based on the linear dimensions of the leaves, making it a nondestructive, economical, and effective technique that allows measurement of the LA throughout the plant growth cycle (Hernández-Fernandéz et al., 2021Hernández-Fernandéz, I. A.; Jarma-Orozco, A.; Pompelli, M. F. Allometric models for non-destructive leaf area measurement of stevia: An in depth and complete analysis. Horticultura Brasileira, v.39, p.205-215, 2021. https://doi.org/10.1590/s0102-0536-20210212
https://doi.org/10.1590/s0102-0536-20210...
).
In studies on fruit trees, the LA indicates the relationship of the leaf surface with the fruit weight, quality, and maturation and estimation of production attributes (Keramatlou et al., 2015Keramatlou, I.; Sharifani, M.; Sabouri, H.; Alizadeh, M.; Kamkar, B. A simple linear model for leaf area estimation in Persian walnut (Juglans regiaL.). Scientia Horticulturae, v.184, p.36-39, 2015. http://dx.doi.org/10.1016/j.scienta.2014.12.017
http://dx.doi.org/10.1016/j.scienta.2014...
). Studies with a focus on LA estimations have been carried out on other fruit trees, such as vines (Teobaldelli et al., 2020Teobaldelli, M.; Rouphael, Y.; Gonnella, M.; Buttaro, D.; Rivera, C. M.; Muganu, M.; Colla, G.; Basile, B. Developing a fast and accurate model to estimate allometrically the total shoot leaf area in grapevines. Scientia Horticulturae , v.259, p.1-9, 2020. https://doi.org/10.1016/j.scienta.2019.108794
https://doi.org/10.1016/j.scienta.2019.1...
) and cashew trees (Gomes et al., 2020Gomes, R. G.; Silva, D. F. P. da; Ragagnin, A. L. S. L.; Souza, P. H. M. de; Cruz, S. C. S. Leaf area estimation of Anacardium humile. Revista Brasileira de Fruticultura, v.42, p.1-8, 2020. http://dx.doi.org/10.1590/0100-29452020628
http://dx.doi.org/10.1590/0100-294520206...
).
Although LA estimation through regression models is easy to apply, there are still no studies on the application of this indirect method for sapodilla crops. Regression models may provide information for subsequent studies on this species. Therefore, the objective of this study was to build allometric equations that allow for the fast and accurate estimation of the sapodilla LA.
# Material and Methods
This study was performed using sapodilla trees grown in the didactic orchard of the Center of Agrarian Sciences, Federal Rural University of the Semi-Arid Region, Rio Grande do Norte, Brazil (5° 11’ S, 37° 20’ W). The climate in this region is classified as BSh and is considered dry and very hot, with a dry season and summer rain (Alvares et al., 2013Alvares, C. A.; Stape, J. L.; Sentelhas, P. C.; Gonçalves, J. L. de M.; Sparovek, G. Köppen’s climate classification map for Brazil. Meteorologische Zeitschrift, v.22, p.711-728, 2013. https://doi.org/10.1127/0941-2948/2013/0507
https://doi.org/10.1127/0941-2948/2013/0...
). The average temperature is approximately 28 °C, and annual rainfall is approximately 695 mm. The soil in the experimental area is classified as Eutrophic Red-Yellow Argisol (Ultisol).
In total, 250 leaves were randomly collected from 15 matrices of M. zapota trees. Expanded, healthy, pest-free, and disease-free leaves or leaves with damage caused by biotic or abiotic related factors were selected (Ribeiro et al., 2022aRibeiro, J. E. da S.; Nóbrega, J. S.; Coêlho, E. dos S.; Dias, T. J.; Melo, M. F. Estimating leaf area of basil cultivars through linear dimensions of leaves. Ceres, v.69, p.139-147, 2022a. http://dx.doi.org/10.1590/0034-737X202269020003
http://dx.doi.org/10.1590/0034-737X20226...
). Leaves of different shapes and sizes were selected to test the generality of the model and to obtain significantly more variability in the sample data. Immediately after collection, the leaf samples were packed in plastic bags and kept in the shade to maintain their turgidity.
The leaves were digitized using a table scanner (model OKI ES5162LP MFP) with a 1,200 × 1,200 DPI resolution. The images were processed, contrasted, and analyzed individually using ImageJ software (National Institutes of Health, USA) according to the methodology described by Ribeiro et al. (2018Ribeiro, J. E. da S.; Barbosa, A. J. S.; Albuquerque, M. B. de. Leaf area estimate of Erythroxylum simonis Plowman by linear dimensions. Floresta e Ambiente, v.25, p.243-250, 2018. https://doi.org/10.1590/2179-8087.010817
https://doi.org/10.1590/2179-8087.010817...
). During the scanning of the images, rules graduated in centimeters were included on each sheet as indicators of the reference scale for the analyses. For each leaf, the maximum length (L) (cm) (distance between the insertion end of the petiole and the opposite distance of the central rib) and maximum width (W) (cm) (superior measure perpendicular to the central rib) were measured for calculation of the actual (observed) LA (cm²) (Figure 1). Then, using the length and width data, the product of length and width (LW) (cm²), product of length and length (LL) (cm²), and product of width and width (WW) (cm²) were calculated.
Figure 1
Length (L) and width (W) of a representative leaf of sapodilla
On the basis of the variance inflation factor (VIF) (Eq. 1) (Marquaridt, 1970Marquaridt, D. W. Generalized inverse, ridge regression, biased linear estimation and nonlinear estimation. Technometrics, v.12, p.591-612, 1970. https://doi.org/10.1080/00401706.1970.10488699
https://doi.org/10.1080/00401706.1970.10...
) and tolerance value (T) (Eq. 2) (Gill, 1986Gill, J. L. Outliers, residuals, and influence in multiple regression. Journal of Animal Breeding and Genetics, v.103, p.161-175, 1986. https://doi.org/10.1111/j.1439-0388.1986.tb00079.x
https://doi.org/10.1111/j.1439-0388.1986...
), the degree of collinearity between the linear parameters of the leaves (L and W) was evaluated to verify the accuracy of the regression coefficient estimates. A variance inflation factor of greater than 10 and a tolerance value of less than 0.1 indicate that the length (L) and width (W) data have multicollinearity, which may affect the estimation of the LA; therefore, one of these parameters should be excluded for the adjustment of the regression models for LA estimation (Gill, 1986).
$V I F = 1 1 - r 2$ (1)
$T = 1 V I F$ (2)
where:
r - correlation coefficient between L and W;
VIF - variance inflation factor; and,
T - tolerance value.
Tests of 90 linear and nonlinear regression models were performed to estimate the LA (dependent variable) as a function of the linear dimensions of the leaves (L, W, LW, LL, and WW) (independent variables). Logarithmic and polynomial models from the second to fifth order were excluded, leaving 16 models that presented satisfactory criteria in addition to high accuracy and speed of analysis.
The regression models tested were as follows: linear (ŷ = β0 + β1 × x + εi); linear without an intercept (ŷ = β1 × x + εi); power (ŷ = β0 × xβ1 + εi); and exponential (ŷ = β0 × β1 x + εi). The value of ŷ estimates the LA as a function of x, which corresponds to the linear dimensions of the leaves (L, W, LW, LL, and WW). Student’s t-test (with a probability value of 0.05) was used to test the following H0 hypotheses: β0 = 0 versus Ha: β0 ≠ 0; and H0: β1 = 0 versus Ha: β1 ≠ 1, where β0 and β1 are regression coefficients.
The best models for estimating the LA of sapodilla were chosen on the basis of the following criteria: a higher coefficient of determination (R²) (Eq. 3), a higher Pearson linear correlation coefficient (r) (Eq. 4), a lower Akaike information criterion (AIC) (Eq. 5), a higher Willmott agreement index (d) (Eq. 6), and a lower root of the mean error square (RMSE) (Eq. 7).
$R 2 = 1 - ∑ i = 1 n y i - y ^ i 2 ∑ i = 1 n y i '$ (3)
$r = ∑ i = 1 n y i - y ¯ x i - x ¯ ∑ i = 1 n y i - y ¯ 2 ∑ i = 1 n x i - x ¯ 2$ (4)
$A I C = - 2 ln L x \ θ ^ + 2 p$ (5)
$d = 1 - ∑ i = 1 n y ^ i - y i 2 ∑ i = 1 n y ^ i ' + y i ' 2$ (6)
$R M S E = ∑ i = 1 n y ^ i - y i 2 n$ (7)
where:
R² - coefficient of determination;
r - Pearson linear correlation coefficient;
AIC - Akaike information criterion;
d - Willmott agreement index;
RMSE - root of the mean error square;
ŷi - estimated values of leaf area;
yi - observed values of leaf area;
yi - average of the observed values;
y’i = ŷi - y; y’i = yi - y; L(x\θ) - maximum likelihood function;
p - number of model parameters;
n - number of observations;
xi and yi - i-th observations of the variables y and x; and,
y and x - means of variables y and x.
Descriptive analyses were performed to calculate the minimum, maximum, and average values; amplitude; standard deviation; standard error; asymmetry; shortness; and coefficient of variation. Data normality was verified using the Shapiro-Wilk test (Shapiro & Wilk, 1965Shapiro, S. S.; Wilk, M. B. Analysis of variance test for normality (complete samples). Biometrika, v.52, p.591-611, 1965.). The observed and estimated LAs were compared using Student’s t-test for paired samples (p < 0.01). Statistical analyses of the data were performed using R® v.4.1.2 (R Core Team, 2022R Core Team. R: A language and environment for statistical computing. Vienna: R Foundation for Statistical Computing, 2022.).
# Results and Discussion
The sapodilla leaves varied in length of between 3.70 and 16.25 cm, with an average of 9.55 cm and amplitude of 12.55 cm (Figure 2A). The leaf width varied between 1.58 and 5.94 cm, with a mean of 3.42 cm and an amplitude of 4.36 cm (Figure 2B). The LW values ranged from 5.87 to 94.85 cm², with an average of 34.41 cm² and amplitude of 88.98 cm² (Figure 2C). The LL values varied between 13.72 and 264.19 cm², with a mean of 96.54 and an amplitude of 250.47 cm² (Figure 2D). The WW values ranged from 2.51 to 35.29 cm², with an average of 12.42 cm² and amplitude of 32.78 cm² (Figure 2E). The actual LA values ranged between 4.22 and 70.86 cm², with an average of 24.47 cm² and amplitude of 66.64 cm² (Figure 2F). The lowest coefficients of variation were observed for leaf length and width, 24.37 and 24.90%, respectively (Figures 2A and B). The highest data variability was recorded for the LW (47.53%), LL (47.92%), WW (50.15%), and actual LA (48.34%) values (Figures 2C, D, E and F).
Figure 2
Descriptive analysis of the length (A), width (B), product of length and width (LW) (C), product of length and length (LL) (D), product of width and width (WW) (E), and leaf area (LA) (F) of 250 sapodilla leaves
A wide variability in LW, LL, WW, and LA data is of fundamental importance for studies involving regression models for estimating the LA of fruit species (Oliveira et al., 2017Oliveira, P. S.; Silva, W.; Costa, A. A. M.; Schmildt, E. R.; Vitória, E. L. da. Leaf area estimation in litchi by means of allometric relationships. Revista Brasileira de Fruticultura , v.39, p.1-6, 2017. https://doi.org/10.1590/0100-29452017403
https://doi.org/10.1590/0100-29452017403...
; Gomes et al., 2020Gomes, R. G.; Silva, D. F. P. da; Ragagnin, A. L. S. L.; Souza, P. H. M. de; Cruz, S. C. S. Leaf area estimation of Anacardium humile. Revista Brasileira de Fruticultura, v.42, p.1-8, 2020. http://dx.doi.org/10.1590/0100-29452020628
http://dx.doi.org/10.1590/0100-294520206...
). High data variability allows for the construction of more representative models and precise equations that can be applied for leaves of different shapes and sizes, which can be measured at different phenological stages during the plant life cycle (Cargnelutti Filho et al., 2021Cargnelutti Filho, A.; Pezzini, R. V.; Neu, I. M. M.; Dumke, G. E. Estimation of buckwheat leaf area by leaf dimensions. Semina Ciências Agrárias, v.42, p.1529-1548, 2021. https://doi.org/10.5433/1679-0359.2021v42n3Supl1p1529
https://doi.org/10.5433/1679-0359.2021v4...
). The collection of a large number of leaves (250 leaves) from different parts of the matrices was considered to be ideal for constructing models that estimate the LA of sapodilla as a function of linear measurements of leaves. Previous studies have confirmed that the use of a low number of samples to build allometric models can lead to the generation of biased equations with low reliability for estimating the LA (Antunes et al., 2008Antunes, W. C.; Pompelli, M. F.; Carretero, D. M.; DaMatta, F. M. Allometric models for non-destructive leaf area estimation in coffee (Coffea arabica and Coffea canephora). Annals of Applied Biology, v.153, p.33-40, 2008. https://doi.org/10.1111/j.1744-7348.2008.00235.x
https://doi.org/10.1111/j.1744-7348.2008...
; Pompelli et al., 2012Pompelli, M. F.; Antunes, W. C.; Ferreira, D. T. R. G.; Cavalcante, P. P. G. S.; Wanderley-Filho, H. C. L.; Endres, L. Allometric models for non-destructive leaf area estimation of the Jatropha curcas. Biomass and Bioenergy, v.36, p.77-85, 2012. https://doi.org/10.1016/j.biombioe.2011.10.010
https://doi.org/10.1016/j.biombioe.2011....
).
The kurtosis coefficients (k) of LW, LL, WW, and LA presented a platykurtic distribution, which was flatter than the normal distribution (k > 3.26). By contrast, the length data presented a leptokurtic distribution (k < 3.26), whereas the width data exhibited a mesokurtic one (k = 3.26) (Figure 2). The high p-values of the normality test (p ≥ 0.05), combined with the magnitude of the mean about the median and asymmetry, characterized a suitable adjustment of the length and width data to the normal distribution. The asymmetry of the LW, LL, WW, and LA data indicated a higher frequency of leaves with values approaching the minimum and a lower frequency of those with values close to the maximum, confirming the non-normality of the data (Ribeiro et al., 2022bRibeiro, J. E. da S.; Figueiredo, F. R. A.; Nóbrega, J. S.; Coêlho, E. dos S.; Melo, M. F. Leaf area of Erythrina velutina Willd. (Fabaceae) through allometric equations. Revista Floresta, v.52, p.93-102, 2022b. http://dx.doi.org/10.5380/rf.v52i1.78059
http://dx.doi.org/10.5380/rf.v52i1.78059...
). Linear and nonlinear association patterns between the L, W, LW, LL, WW, and LA values were observed in the dataset used to construct the predicted regression models for LA estimation (Figure 3). Linear patterns were observed between LW and LA, LL and LA, and WW and LA, whereas nonlinear patterns were evident between L and LA and W and LA (Figure 3), indicating the need for different regression models for data adjustment and validation.
Figure 3
Frequency histograms (diagonal) and data dispersion between the length, width, product of length and width, product of length and length, product of width and width, and leaf area of 250 sapodilla leaves used to build equations for estimating the leaf area
The variance inflation factor (VIF) values ranged between 0.004 and 0.139, whereas the tolerance (T) values ranged from 7.165 to 217.64. Thus, for all constructed models, the variance inflation factor values were less than 10 and the tolerance values were greater than 0.10, indicating that the collinearity between the length and width data was negligible and these parameters were valid for use in the regression models (Gill, 1986Gill, J. L. Outliers, residuals, and influence in multiple regression. Journal of Animal Breeding and Genetics, v.103, p.161-175, 1986. https://doi.org/10.1111/j.1439-0388.1986.tb00079.x
https://doi.org/10.1111/j.1439-0388.1986...
; Fanourakis et al., 2021Fanourakis, D.; Kazakos, F.; Nektarios, P. A. Allometric individual leaf area estimation in chrysanthemum. Agronomy, v.11, p.795, 2021. https://doi.org/10.3390/agronomy11040795
https://doi.org/10.3390/agronomy11040795...
).
The models presented coefficient of determination (R²) values of above 0.86, indicating that at least 86% of the variations in sapodilla LA were explained by the equations proposed for the estimation (Table 1). The equations that used the LW value presented the best criteria for estimating the LA of the species, giving the best fits of the regression models (Macário et al., 2020Macário, A. P. S.; Ferraz, R. L. de S.; Costa, P. da S.; Brito Neto, J. F. de; Melo, A. S. de; Dantas Neto, J. Allometric models of estimating Moringa oleífera leaflets area. Ciência e Agrotecnologia, v.44, p.1-10, 2020. https://doi.org/10.1590/1413-7054202044005220
https://doi.org/10.1590/1413-70542020440...
; Goergen et al., 2021Goergen, P. C. H.; Lago, I.; Schwab, N. T.; Alves, A. F.; Freitas, C. P. de O.; Verlaine, S.; Selli, V. S. Allometric relationship and leaf area modeling estimation on chia by non-destructive method. Revista Brasileira de Engenharia Agrícola e Ambiental, v.25, p.305-311, 2021. https://doi.org/10.1590/1807-1929/agriambi.v25n5p305-311
https://doi.org/10.1590/1807-1929/agriam...
). The exception was the exponential model, where the best criteria were observed in the equation in which the WW value was used (Ribeiro et al., 2020Ribeiro, J. E. da S.; Coêlho, E. dos S.; Figueiredo, F. R. A.; Melo, M. F. Non-destructive method for estimating leaf area of Erythroxylum pauferrense (Erythroxylaceae) from linear dimensions of leaf blades. Acta Botanica Mexicana, v.127, p.1-12, 2020. https://doi.org/10.21829/abm127.2020.1717
https://doi.org/10.21829/abm127.2020.171...
).
Table 1
Models, regression coefficients (β0 and β1), coefficient of determination (R²), Pearson’s linear correlation coefficient (r), Akaike information criterion (AIC), Willmott agreement index (d), root mean square error (RMSE), and equations for estimating the leaf area (LA) of sapodilla as a function of linear leaf dimensions (length and width)
The criteria used to choose the best equations for estimating the LA of sapodilla through linear dimensions of the leaves confirmed that the power model and the linear model without the intercept, both constructed using the LW values, were the quickest and most accurate. These models showed the highest coefficients of determination (R²) (0.9991 and 0.9954), Pearson’s linear correlation coefficients (r) (0.9977 and 0.9976), and Willmott agreement indexes (d) (0.9988 and 0.9988) and the lowest Akaike information criterion (AIC) (560.75 and 567.08) and root mean square error values (RMSE) (0.7998 and 0.8108) (Table 1). Linear and power models were also the most suitable for estimating the LA of other plant species (Tondjo et al., 2015Tondjo, K.; Brancheriau, L.; Sabatier, S.; Kokutse, A. D.; Akossou, A.; Kokou, K.; Fourcaud, T. Non-destructive measurement of leaf area and dry biomass in Tectona grandis. Trees, v.29, p.1625-1631, 2015. https://dx.doi.org/10.1007/s00468-015-1227-y
https://dx.doi.org/10.1007/s00468-015-12...
; Trachta et al., 2020Trachta, M. A.; Zanon, A. J.; Alves, A. F.; Freitas, C. P. O.; Streck, N. A.; Cardoso, P. S.; Rodrigues, L. B. Leaf area estimation with nondestructive method in cassava. Bragantia, p.79, p.472-484, 2020. https://doi.org/10.1590/1678-4499.20200018
https://doi.org/10.1590/1678-4499.202000...
; Montelatto et al., 2021Montelatto, M. B.; Villamagua-Vergara, G. C.; Brito, C. M. de; Castanho, F.; Sartori, M. M.; Silva, M. de A.; Guerra, S. P. S. Bambusa vulgaris leaf area estimation on short-rotation coppice. Scientia Forestalis, v.49, p.1-9, 2021. https://doi.org/10.18671/scifor.v49n129.14
https://doi.org/10.18671/scifor.v49n129....
; Mela et al., 2022Mela, D.; Dias, M. G.; Silva, T. I. da; Ribeiro, J. E. da S.; Martinez, A. C. P.; Zuin, A. H. L. Estimation of Thunbergia grandiflora leaf area from allometric models. Comunicata Scientiae, v.13, p.1-6, 2022. https://doi.org/10.14295/cs.v13.3722
https://doi.org/10.14295/cs.v13.3722...
).
The equations proposed for estimating the LA of sapodilla presented high adjustments of the data (R² > 0.99), in which the residual variance was homogeneous, with little dispersion of the data (Figure 4). The LA data estimated from the constructed equations showed positive correlations with the observed values (measured from the digital images), producing coefficient of determination (R²) values of greater than 0.99 (Figures 5A and C). There were no significant differences between the observed LAs and the values estimated in the indicated models, confirming the significant relationship between the observed and estimated data (Figures 5B and D). Thus, the equations ŷ = 0.664 × LW1.018 (power model) and ŷ = 0.713 × LW (linear model without intercept) are the most suitable for accurately estimating the LA (> 99%) of sapodilla by means of the linear dimensions of the leaf. These two equations can also be used to measure this parameter in different cultivation environments and phenological stages of the species.
Figure 4
Relationship between the observed leaf area and the product of length and width of sapodilla leaves calculated using the models ŷ = 0.664 × LW1.018 and ŷ = 0.713 × LW. An analysis of the dispersion pattern of the residues is presented in the inset
Figure 5
Relationship and comparison (Student’s t-test) between the observed leaf area and the leaf area estimated using the linear model without intercept (A and B) and the power model (C and D) as a function of the product length and width of the sapodilla leaves
# Conclusions
1. The leaf area of sapodilla can be estimated with a nondestructive indirect method using allometric equations based on the linear dimensions of the leaves.
2. The equations ŷ = 0.664 × LW1.018 (R² = 0.9991) and ŷ = 0.713 × LW (R² = 0.9954), which use the product of the leaf length and width, are the most suitable for the accurate and quick estimation of the leaf area of sapodilla.
# Literature Cited
• Alvares, C. A.; Stape, J. L.; Sentelhas, P. C.; Gonçalves, J. L. de M.; Sparovek, G. Köppen’s climate classification map for Brazil. Meteorologische Zeitschrift, v.22, p.711-728, 2013. https://doi.org/10.1127/0941-2948/2013/0507
» https://doi.org/10.1127/0941-2948/2013/0507
• Antunes, W. C.; Pompelli, M. F.; Carretero, D. M.; DaMatta, F. M. Allometric models for non-destructive leaf area estimation in coffee (Coffea arabica and Coffea canephora). Annals of Applied Biology, v.153, p.33-40, 2008. https://doi.org/10.1111/j.1744-7348.2008.00235.x
» https://doi.org/10.1111/j.1744-7348.2008.00235.x
• Cargnelutti Filho, A.; Pezzini, R. V.; Neu, I. M. M.; Dumke, G. E. Estimation of buckwheat leaf area by leaf dimensions. Semina Ciências Agrárias, v.42, p.1529-1548, 2021. https://doi.org/10.5433/1679-0359.2021v42n3Supl1p1529
» https://doi.org/10.5433/1679-0359.2021v42n3Supl1p1529
• Fanourakis, D.; Kazakos, F.; Nektarios, P. A. Allometric individual leaf area estimation in chrysanthemum. Agronomy, v.11, p.795, 2021. https://doi.org/10.3390/agronomy11040795
» https://doi.org/10.3390/agronomy11040795
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» https://doi.org/10.1111/j.1439-0388.1986.tb00079.x
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» https://doi.org/10.1371/journal.pone.0235499
• 1 Research developed at Universidade Federal Rural do Semi-Árido, Mossoró, RN, Brazil
### Edited by
Editors: Ítalo Herbet Lucena Cavalcante & Hans Raj Gheyi
# Publication Dates
• Publication in this collection
21 Nov 2022
• Date of issue
Mar 2023
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2023-02-06 22:49:36
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http://www.r-bloggers.com/the-psychology-of-music-and-the-%E2%80%98tuner%E2%80%99-package/
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# The Psychology of Music and the ‘tuneR’ Package
October 25, 2011
By
(This article was first published on John Myles White » Statistics, and kindly contributed to R-bloggers)
### Introduction
This semester I’m TA’ing a course on the Psychology of Music taught by Phil Johnson-Laird. It’s been a great course to teach because (i) so much of the material is new to me and (ii) because the study of the psychology of music brings together so many of the intellectual tools I enjoy, including music theory, psychophysics and Fourier analysis.
One topic this semester that was completely new to me was the theory of tuning: I had known about the invention of the well-tempered system of tuning, but had never heard of Pythagorean tuning or just tuning — and certainly was not aware that the well-tempered system Bach celebrated was not identical to our current equal-tempered system of tuning.
As a way of consolidating some of the knowledge I’ve gained, I decided I’d write a blog entry after several months of neglecting this blog. (For that neglect, I’ll blame a combination of grant writing, book writing, ongoing research projects and personal life developments.) In what follows, I’ll give a brief overview of the theory of tuning at a theoretical level that should be accessible to anyone who’s familiar with the names of intervals and feels comfortable thinking quantitatively.
After surveying the field, I’ll turn to a discussion of some code I’ve written in R that implements these ideas using the ‘tuneR’ package, which is one of my favorite hidden gems from CRAN. Along the way, I’ll introduce some of the simplest tools from the ‘tuneR’ package that can be used for generating computer music.
### Tuning Systems: Pythagorean, Just and 12-Tet
It’s worth noting right at the start that tuning is a misleading name for the topic we’ll be discussing: we’re not talking about how one tunes a fixed instrument so that it sounds in tune, but rather we’re interested in how one defines the very notes that the instrument should be able to produce when it’s perfectly in tune.
To make that clear, let’s assume that we’ve accepted as a given that a frequency of 440 Hz will be called A. Our problem then becomes one of deciding which of the infinitely many frequencies we could produce actually deserves the label of A#, B, C, C#, and so on.
#### Pythagorean Tuning
The simplest solution to this problem I know of is the Pythagorean tuning system. It’s based on constructing all of the possible notes using a series of perfect fifths. If you remember the Circle of Fifths, you’ll remember that you can reach every chromatic note by ascending fifths: if you start at A, you’ll proceed through E, B, F# and so on.
The Pythagorean system implements the Circle of Fifths directly using repeated multiplication of a base frequency. To do this, you first declare that a perfect fifth is at a frequency 3/2 above your base frequency. For example, this definition implies that the perfect fifth above the A at 440 Hz has to be at a frequency of 3/2 * 440 = 660 Hz. Once you do this, you’ve defined the frequency we’ll call E.
And following on with this logic, you produce a B at 990 Hz. Of course, this B occurs an octave above the base A at 440 Hz, so you transpose it down an octave to produce the B you’ll actually use. To do this, you need to assume that an octave is at a frequency 2 times the base frequency. Since we’ve accepted that 990 Hz is a B, we divide 990 by 2 and conclude that 495 Hz should be B.
With these three notes defined, we have the following table of frequency/note pairs:
Note Frequency Ratio with 440 Hz
A 440 Hz 1
E 660 Hz 3/2
B 495 Hz 9/8
If we continue on with this logic and calculate many more multiplications by 3/2 and divisions by 2, we will eventually produce a complete table for all of the notes in the chromatic scale that looks like the following:
Note Frequency Ratio
A 440 1
A# 463.5391 256/243
B 495 9/8
C 521.4815 32/27
C# 556.875 81/64
D 586.6667 4/3
D# 626.4844 729/512
E 660 3/2
F 695.3086 128/81
F# 742.5 27/16
G 782.2222 16/9
G# 835.3125 243/128
A 880 2
One thing about this table might strike you as odd if you’re mathematically savvy: the octave, which we’ve defined by fiat as a ratio of 2:1, could never have been produced by successive multiplication by 3/2, since no power of 3 will be evenly divisible by a power of 2. This is the one flub in the Pythagorean system: you can’t really produce the entire chromatic scale using only multiples of 3/2. Here we’ve solved that problem by replacing the note we would have called A with a true octave generated using multiplication by 2. Because the exact octave produced by Pythagorean tuning is slightly out of tune with our preferred definition of an octave, you may hear people refer to this discrepancy as the the Pythagorean comma.
#### Just Tuning
Given that we had to cheat a bit to create a proper octave using the Pythagorean tuning system based on multiples of 3/2, it makes sense to ask why we shouldn’t just allow ourselves to use other multipliers than 3/2. Looking at the Pythagoren tuning table, we see some pretty ugly fractions like 729/512. What if we forced these fractions to be simpler by employing ratios like 4/3 and 5/4 to build up the whole system?
The result of allowing ourselves several fractions beyond just those derived from 3/2 is called the just tuning system. Here we assume that perfect fifths occur at a frequency ratio of 3/2 and that perfect fourths occur at a frequency ratio of 4/3. Continuing on with this process, we eventually end up with the following tuning table:
Note Frequency Ratio
A 440 1
A# 469.3333 16/15
B 495 9/8
C 528 6/5
C# 550 5/4
D 586.6667 4/3
D# 625.7778 64/45
E 660 3/2
F 704 8/5
F# 733.3333 5/3
G 782.2222 16/9
G# 825 15/8
A 880 2
This is the tuning that early Classical music was written in. Looking at the table you con immediately appreciate the theoretical assertion that the relative dissonance of an interval is determined by the simplicity of the ratio of frequencies between the two notes: perfect fifths are 3/2 and major thirds are 5/4, while minor seconds are 16/15 and major sevenths are 15/8. This is one of the things I most enjoy about the theory of harmony: there’s a match between the aesthetics of fractions and the aesthetics of sounds that, for me, helps to justify my sense that certain fractions are more beautiful than others.
#### 12 Tet / Equal-Temperament
Now, if you know the history of Bach’s Well-Tempered Clavier, you know that there is a problem with the just tuning system: it sounds great in the key you used as the base (here A), but it sounds a bit out of tune in other keys. The modern 12-tet system is the most recent approach to solving this problem: you assume the gap between two semitones (e.g. A to A# or A# to B) is always the exact same multiple. Since you’ll repeat this multiplication 12 times before reaching an octave, you can conclude that two notes that are a semitone apart must be separated by the 12th root of 2. Building a tuning system using that ratio alone gives us our modern system of tuning, which is shown in the table above using the decimal expansion of the ratios instead of their representation as powers of the 12th root of 2:
Note Frequency Ratio
A 440 1.000000
A# 466.1638 1.059463
B 493.8833 1.122462
C 523.2511 1.189207
C# 554.3653 1.259921
D 587.3295 1.334840
D# 622.2540 1.414214
E 659.2551 1.498307
F 698.4565 1.587401
F# 739.9888 1.681793
G 783.9909 1.781797
G# 830.6094 1.887749
A 880 2.000000
### Listening to the Results
We’ve just described three ways to define the notes used in Western music. But how different do they sound? To answer that, I decided to produce a series of simple sine wave audio samples that were tuned using each of the three tuning systems. To produce those audio samples, I used the ‘tuneR’ package, which I’ll describe now. Before you read on, you should install it from CRAN using the standard install.packages('tuneR') invocation.
### A tuneR Tutorial
The tuneR package is an extremely convenient tool for generating audio files from R based on a numeric description of the audio stream. For the purposes of this discussion of tuning systems, we simply need to produce basic sine waves. Thankfully, that’s very easy to do with tuneR. Here’s an example:
1 2 3 4 5 library('tuneR') sound <- sine(440, bit = 16) writeWave(sound, '440.wav')
Here we’ve loaded the tuneR package, created a 1s snippet of sine wave audio at 16 bits resolution using the sine function, and then written out the audio to a WAV file using writeWave. If you look at your current directory and listen to this file, you’ll hear a sine wave at 440 Hz.
If you want to explore the use of sine, you can easily play with the duration of the sound by changing the duration parameter. If you want to, you can also change the sample rate and the bit rate, but I don’t see any reason to do that while exploring ideas about tuning.
More important is knowing that you can superimpose two sine waves using the + operator and that you can concatenate them using the bind function. To show off producing octaves, for example, you might use the following code to hear an A at 440 Hz, then an A an octave above it, and finally the harmony they produce together:
1 2 3 4 5 6 7 library('tuneR') sound <- bind(sine(440, bit = 16), sine(880, bit = 16), sine(440, bit = 16) + sine(880, bit = 16)) writeWave(sound, 'octaves.wav')
Unfortunately, this sample code produces an error because of the naive addition we’ve implemented using the + operator. Adding two sine waves directly together overfills the bit rate we’re using. To safely perform addition of two sine waves, we need to normalize the results of our summation using the normalize function. This gives us just one more line of code:
1 2 3 4 5 6 7 8 9 library('tuneR') sound <- bind(sine(440, bit = 16), sine(880, bit = 16), sine(440, bit = 16) + sine(880, bit = 16)) sound <- normalize(sound, unit = '16') writeWave(sound, 'octaves.wav')
For reasons that are not clear to me, you have to specify the bit rate to normalize using the unit parameter rather than the bit parameter.
### Demoing Tuning Systems
Our little octave demo is cute, but we really want to know what more interesting harmonies like major thirds and minor seconds sound like in the various tuning systems we described. To do that, I first wrote a function called interval that spits out the multiplier you need to use to produce a given interval for any of the three tuning systems. That function is in a GitHub repository I’ve set up with code for making these demos. If you download that repository, you could load my interval function using a simple call to source like the one seen below. And using this interval function, we can generate demos of various intervals as follows:
1 2 3 4 5 6 7 8 9 10 11 library('tuneR') source('interval.R') base <- 440 sound <- sine(base) + sine(interval('minor-second', tuning = 'pythagorean') * base) sound <- normalize(sound, unit = '16') writeWave(sound, 'minor_second_pythagorean.wav')
On GitHub there’s a file called test_intervals.R that will go through and generate all of the intervals in all three tuning systems. If you run that file, you’ll generate a lot of audio files you can listen to as demos of the three tuning systems we’ve described. For me, these tuning systems all produce intervals that sound surprisingly similar, though at high volumes I find it moderately easy to hear slight differences between the tuning systems. That said, I very much doubt I would pick up on them in a normal musical context.
That’s the end of my little introduction to tuning systems and the use of the tuneR package to explore them. If you’re interested in thinking computationally about music, I highly recommend playing around with tuneR until you feel like you can produce interesting results. I’m already working on trying to build up some interesting timbres to work with.
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2014-11-01 08:26:41
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https://www.intechopen.com/books/automation-in-agriculture-securing-food-supplies-for-future-generations/robotic-harvesting-of-fruiting-vegetables-a-simulation-approach-in-v-rep-ros-and-matlab/
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Engineering » Electrical and Electronic Engineering » "Automation in Agriculture - Securing Food Supplies for Future Generations", book edited by Stephan Hussmann, ISBN 978-953-51-3874-7, Print ISBN 978-953-51-3873-0, Published: March 14, 2018 under CC BY 3.0 license. © The Author(s).
# Robotic Harvesting of Fruiting Vegetables: A Simulation Approach in V-REP, ROS and MATLAB
By Redmond R. Shamshiri, Ibrahim A. Hameed, Manoj Karkee and Cornelia Weltzien
DOI: 10.5772/intechopen.73861
Article top
## Overview
Figure 1. Manual harvesting of fruits.
Figure 2. Research and development in robotic harvesting of fruits with different manipulators and gripper mechanisms for: (A) citrus, (B, C) sweet pepper, (D, E) tomato, (F) cucumber, (G, H) strawberry, and (I–K) apple.
Figure 3. Diagram showing ROS file architecture and nodes communicating system.
Figure 4. Schematic diagram showing the architecture and the main elements of V-REP simulator.
Figure 5. Demonstration of the steps in the robust image processing algorithm using edge detection with fuzzy-logic for identification and tracking of sweet pepper.
Figure 6. Image publishing and subscribing in ROS, Left image: snapshot of the main ROS package folders in the V-REP, and right image: snapshot of the simulation environment in V-REP publishing an image to a ROS node.
Figure 7. Major scene objects used in the simulation.
Figure 8. Major sensors used in the experiments, first row image are the actual and the second row images are the simulated sensors.
Figure 9. Filter used by each of the two vision sensors in “Fast Hokuyo URG-04LX-UG01” V-REP model (top), and by the vision sensor in the “3D laser scanner Fast” V-REP model (bottom).
Figure 10. CAD models of the entire plant system: fruit, leaves, stem, calyx, and leaves.
Figure 11. Simulation scene setup with CAD models of the professional robot manipulator used in visual servo control experiment, left: Fanuc LR Mate 200iD, right: NOVABOT.
Figure 12. Simulation scene setup with CAD models of the multiple linear actuator robotic platform.
Figure 13. Simulation scene setup with CAD models of the multiple SCARA arm robotic platform.
Figure 14. Demonstration of (A) five different control mechanisms in V-REP, (B) inverse kinematics chain, (C, D) minimum distance calculation from tip vision sensor and plant/fruit model, (E) collision detection between robot links and plant model, and (F) path planning for moving a harvested fruit to the bin.
Figure 15. Demonstration of the joint functionality and the inverse kinematics chain for the NOVABOT manipulator.
Figure 16. Visual servo control scheme with eye in hand configuration based on image moment method used with the Fanuc LR Mate 200iD and the NOVABOT manipulators.
Figure 17. Schematic diagram of an innovative approach for robotic harvesting based on multiple low-cost manipulators (e.g., multiple linear actuators or SCARA arms).
Figure 18. Results of the image processing algorithm for fruit localization using wide camera view (left) and divided camera view (right).
Figure 19. Result of the image processing algorithm for quantification and tracking of sweet pepper in different fruit-and-plant scenario.
Figure 20. Two dimensional scanning experiment (x-y, x-z, and y-z planes) for finding the maximum fruit visibility. Camera was set to move at 30 degrees increments around the fruit and plant model.
Figure 21. Three dimensional scanning experiments (x-y-z space) for finding the maximum fruit visibility. Camera was set to rotate around the fruit and plant until it finds the best angle of attack.
Figure 22. Simulation of scanning experiment with Laser scanners and depth sensors.
Figure 23. Simulation of visual servo control experiment with the eye-in-hand configuration and PID Control law on joint angles with feedbacks from image moments. Stability was achieved in 2.5 s without overshoot and oscillations.
Figure 24. Simulation of robotic harvesting with arrays of linear actuators.
# Robotic Harvesting of Fruiting Vegetables: A Simulation Approach in V-REP, ROS and MATLAB
Redmond R. Shamshiri1, 3, Ibrahim A. Hameed1, Manoj Karkee2 and Cornelia Weltzien3
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## Abstract
In modern agriculture, there is a high demand to move from tedious manual harvesting to a continuously automated operation. This chapter reports on designing a simulation and control platform in V-REP, ROS, and MATLAB for experimenting with sensors and manipulators in robotic harvesting of sweet pepper. The objective was to provide a completely simulated environment for improvement of visual servoing task through easy testing and debugging of control algorithms with zero damage risk to the real robot and to the actual equipment. A simulated workspace, including an exact replica of different robot manipulators, sensing mechanisms, and sweet pepper plant, and fruit system was created in V-REP. Image moment method visual servoing with eye-in-hand configuration was implemented in MATLAB, and was tested on four robotic platforms including Fanuc LR Mate 200iD, NOVABOT, multiple linear actuators, and multiple SCARA arms. Data from simulation experiments were used as inputs of the control algorithm in MATLAB, whose outputs were sent back to the simulated workspace and to the actual robots. ROS was used for exchanging data between the simulated environment and the real workspace via its publish-and-subscribe architecture. Results provided a framework for experimenting with different sensing and acting scenarios, and verified the performance functionality of the simulator.
Keywords: agricultural robots, automated harvesting, simulation, visual servo control, image processing
## 1. Introduction
Traditional harvesting of fruiting vegetables for fresh market is a labor-intensive task that demands shifting from tedious manual operation to a continuously automated harvesting. In spite of the advances in agricultural robotics, million tons of fruits and vegetables are still hand-picked every year in open-fields and greenhouses (Figure 1). Other than the high labor cost, the availability of the skilled workforce that accepts repetitive tasks in the harsh field conditions impose uncertainties and timeliness costs. For robotic harvesting to be cost-effective, fruit yield needs to be maximized to compensate the additional automation costs. This leads to growing plants at higher densities which make it even harder for an autonomous robot to simultaneously detect the fruit, localize, and harvest it. In the case of sweet pepper fruit, with an estimated yield of 1.9 million tons/year in Europe, reports indicate that while an average time of 6 s per fruit is required for automated harvesting, the available technology has only achieved a success rate of 33% with an average picking time of 94 s per fruit [1]. For cucumber harvesting, a cycle time of 10 s was proven to be economically feasible [2]. Only in Washington State, 15–18 billion apple fruits are harvested manually every year. An estimated 3 million tons of apples is reported to have been produced in Poland in 2015 [3], out of which one-third are delicate fruits and are less resistant to bruising from mass harvester machines. Also in Florida, where the current marketable yield of sweet pepper fruits in open-field cultivation is 1.6–3.0 with potential yield of 4 lb/ft2 in passive ventilated greenhouses [4], manual harvesting is still the only solution. Therefore, development of an automated robotic harvesting should be considered as an alternative method to address the associated labor shortage costs and timeliness.
#### Figure 1.
Manual harvesting of fruits.
Research and development in agricultural robotics date back to 1980s, with Japan, the Netherlands, and the USA as the pioneer countries. The first studies used simple monochrome cameras for apple detection inside the canopy [5]. Advances in the sensor technology and imaging devices have led to the employment of more sophisticated devices such as infrared [6], thermal [7] and hyperspectral cameras [8], or combination of multi-sensors [9] that are adopted with novel vision-based techniques for extracting spatial information from the images for fruit recognition, localization, and tracking. Examples of some of the recent achievements include automatic fruit recognition based on the fusion of color and 3D feature [10], multi-template matching algorithm [11], and automatic fruit recognition from multiple images [12]. Unlike the industrial case, an agriculture robot has to deal with different arrangement of plantings size and shapes, stems, branches, leaves, fruit color, texture, and different location of fruits and plants with respect to each other. Significant contributions have been made by different research groups to address these challenges; however, there is currently no report of a commercial robotic harvesting for fresh fruit market [13], mainly due to the extremely variable heterogeneous working condition and the complex and unpredicted tasks involved with each scenario. Some of the questions to be addressed in designing of a complete robotic harvesting are the simultaneous localization of fruit and environment mapping, path planning algorithms, and the number of detectable and harvestable fruits in different plant density conditions. The function of a robot can be separated into three main sections as sensing (i.e., fruit recognition), planning (i.e., hand-and-eye coordination), and acting (i.e., end-effector mechanism for fruit grasping) [14]. A common approach in fruit detection is by using a single view point, as in the case of a cucumber harvesting robot [15], or multiple viewpoints with additional sensing from one or few external vision sensors that are not located on the robot [16]. Other than the issues with frame transformation, this solution is not promising if the fruit is heavily occluded by the high density plant leaves [17]. Obviously, the final robot prototype needs to be relatively quicker for mass-harvest, with an affordable cost for greenhouse growers. Swarms of simple robots with multiple low-cost camera and grippers, or human-robot collaboration are the research topics to solve the facing challenges in robotic harvesting that current technology cannot overcome. These approaches can significantly improve the processing time of multiple fruit detection in the high-density plants, and provide ground truth results over time for machine learning algorithms based on human-operators experience. Research on agricultural robotics with a focus on automated harvesting of fruiting and vegetable are huge. See for example the works carried out on sweet pepper [1, 18, 19, 20], oil palm [21], cucumber [15, 22, 23, 24], apple [25], strawberry [26, 27], cherry fruit [6], citrus [28], and tomato [29]. Most of these works have used eye-in-hand look-and-move configuration in their visual servo control (Figure 2). Other researches are concentrated on the end-effector design [30], analysis of the robot performance in the dense obstacle environments using stability tests [31], motion planning algorithms [32], and orchard architecture design for optimal harvesting robot [33]. In addition, several software frameworks have been developed for agricultural robotics. An example includes the work of [34], in which a generic high-level functionality was provided for easier and faster development of agricultural robots. Some of the most recent advances in sensing for robotic harvesting include the works of [29, 35] which address the problem of detecting fruits and obstacles in dense foliage. Moreover, [20] and [25] have extensively explored the use of combined color distance, or RGB-D, data on apples and on sweet-peppers, respectively, while [36] present a study devoted to symmetry analysis in three-dimensional shapes for products detection on the plant.
#### Figure 2.
Research and development in robotic harvesting of fruits with different manipulators and gripper mechanisms for: (A) citrus, (B, C) sweet pepper, (D, E) tomato, (F) cucumber, (G, H) strawberry, and (I–K) apple.
Improvement of robotic harvesting requires experimenting with different sensors and algorithms for fruit detection and localization, and a strategy for finding the collision-free paths to grasp the fruits with minimum control effort. Experiments with the actual hardware setup for this purpose are not always feasible due to time constraints, unavailability of equipment (i.e., sensors, cameras, and the robot manipulator), and the operation costs. In the other hand, some hardware setups may result in actuator saturation, or create unsafe situation to the operators and/or plants system. Simulation offers a reliable approach to bridge the gap between innovative ideas and the laboratory trials, and therefore can accelerate the design of a robust robotic fruit harvesting platform for efficient, cost-effective and bruise-free fruit picking. This research was motivated based on the sensing task in robotic harvesting, which requires delivering a robust pragmatic computer vision package to localize mature pepper fruits and its surrounding obstacles. The main objective was to create a completely simulated environment for improvement of plant/fruit scanning and visual servoing task through an easy testing and debugging of control algorithms with zero damage risk to the real robot and to the actual equipment. The research was carried out in two main phases: (i) the creation of the simulated workspace in the virtual robot experimentation platform (V-REP), and (ii) the development of communication and control architecture using the robot operating system (ROS) and MATLAB (The MathWorks Inc., Natick, MA, USA). The simulated workspace included an exact replica of the Fanuc LR Mate 200iD robot manipulator with six degrees of freedom (Fanuc America Corporation, Rochester Hills, MI), models of sweet pepper fruit and plant system, and different vision sensors were created in (V-REP). A simulated color camera attached to the end-effector of the robot was used as fruit localization sensor. ROS was used for exchanging data between the simulated environment and the real workspace via its publish-and-subscribe architecture. This provides a tool for validating the simulated results with those from experimenting with a real robot. V-REP and MATLAB were also interfaced to create two-way communication architecture for exchanging sensors and robot control messages. Data from the simulated manipulator and sensors in V-REP were used as inputs of a visual servo control algorithm in MATLAB. Results provided a flexible platform that saves in cost and time for experimenting with different control strategies, sensing instrumentation, and algorithms in automated harvesting of sweet pepper.
## 2. Overview of the simulation environment
Computer simulation of a complete robotic harvesting task requires: (i) CAD file setup including good replications of the plants-and-fruit scene and the robot manipulators, (ii) simulation environment and calculation modules for the manipulator candidates and platforms (i.e., inverse kinematics and path planning), (iii) different sensors setup, and more importantly (iv) algorithms for control tasks such as visual servoing and gripper control mechanism. The main simulation environment, scene objects, and calculation modules were built in the latest version of V-REP Pro Edu V3.4.0 for Linux 64 (available at www.coppeliarobotics.com), and ROS installed on Ubuntu 14.04.3 LTS. Some of the used terminal commands are summarized in Table 1.
CommandsDescriptionCommandsDescription
File commandsSystem info
lsDirectory listingdateShow the current date and time
ls -alFormatted listing with hidden filescalShow this month’s calendar
cd dir_nameChange directory to dir_nameuptimeShow current uptime
cd ~Change to homewDisplay who is online
pwdShow current directorywhoamiWho you are logged in as
mkdir dir_nameCreate a directory dir_namefinger userDisplay information about user
rm file_nameDelete fileuname -aShow kernel information
rm -r dir_nameDelete directory dir_namecat /proc./cpuinfoCPU information
rm -f file_nameForce remove filecat /proc./meminfoMemory information
rm -rf dir_nameForce remove directory dir_nameman commandShow the manual for command
cp file_name_1 file_name_2Copy file1 to file2dfShow disk usage
cp -r dir_name1 dir_name2Copy dir1 to dir2;duShow directory space usage
mv file_name_1 file_name_2Rename or move file_name_1 to file_name_2
Working with compressed filesShortcuts
tar xf file.tarExtract the files from file.tarCtrl + Alt + TOpens a new terminal window:
tar czf file.tar.gz filesCreate a tar with gzip compressionShift + Ctrl + TOpens a new terminal tab:
tar xzf file.tar.gzExtract a tar using gzipCtrl+CHalts the current command
tar cjf file.tar.bz2Create a tar with bzip2 compressionCtrl+ZStops the current command,
tar xjf file.tar.bz2Extract a tar using bzip2Ctrl+DLog out of current session, exit
gzip fileCompresses file and renames it to file.gzCtrl+WErases one word in the current line
gzip -d file.gzDecompresses file.gz back to fileCtrl+UErases the whole line
Install from sourceCtrl+RType to bring up a recent command
./configure(For example, ./vrep.sh will run v-rep)!!Repeats the last command
dpkg -i pkg.debInstall a package (debian)exitLog out of current session
rpm -Uvh pkg.rpmInstall a package (rpm)
#### Table 1.
List of the most used Ubuntu terminal commands used for navigating in the simulation environment.
ROS Indigo was used to provide a bidirectional communication (information exchange) between simulated robot and cameras with the real world. Experimental packages for Fanuc manipulators within ROS-Industrial (available at http://wiki.ros.org/fanuc_experimental) were used for controlling the manipulator. This design allows reading information from the simulation scene (i.e., robot joints velocity, position, sensors, etc.) and publishes them across ROS network for further process. Results can be used by the simulation, and/or by the real robots and controllers. The image-based visual servo control was carried out in V-REP and MATLAB. For the sake of this chapter, we only provide a brief description of ROS and V-REP.
ROS is a collection of software frameworks for robot software development. It was originally developed in 2007 by the Stanford Artificial Intelligence Laboratory, and with the support of the Stanford AI Robot project. It provides a solution to specific set of problems encountered in the developing large-scale service robots, with philosophical goals summarized as: (i) peer-to-peer, (ii) tools-based, (iii) multi-lingual, (iv) thin, and (v) free and open-source [37]. From 2008 until 2013, development was performed primarily at Willow Garage, a robotics research institute/incubator. During that time, researchers at more than 20 institutions collaborated with Willow Garage engineers in a federated development model. Since 2010, ROS has released several versions, including Box Turtle (March, 2010), C Turtle (August, 2010), Diamondback (March, 2011), Electric Emys (August, 2011), Fuerte Turtle (April, 2012), Groovy Galapagos (December, 2012), Hydro (September, 2013), Indigo (July, 2014), and Jade (May, 2015). The open-source ROS makes it possible to develop code and applications that can be shared and used in other robotic system with minimum effort. It also offers standard operating system features such as hardware abstraction, low-level device control, implementation of commonly used functionalities, message passing between processes, and package management. ROS Packages are files and folders that are built to create minimal collections of code for easy reuse. A ROS package usually includes the followings folders and files: bin, msg, scripts, src, srv, CMakeLists.txt, manifest.xml (Figure 3).
#### Figure 3.
Diagram showing ROS file architecture and nodes communicating system.
Fundamental concepts of the ROS are: Nodes, Messages, Topics, and Services. ROS works based on a “publish-and-subscribe” architecture where processes (called nodes) publish and/or subscribe to specific topic on which information is exchanged in the form of messages (Figure 3). A Node is an executable file that uses ROS to communicate with other Nodes. A Message is ROS data type that is used when subscribing or publishing to a topic. Nodes can publish messages to a Topic as well as subscribe to a Topic to receive messages. Service helps Nodes find each other. ROS nodes use a ROS client library to communicate with other nodes. Nodes can also provide or use a Service. With this architecture, each node in ROS is able to respond to input and activate other nodes, allowing participation of a sequence of nodes to complete complicated robot mission tasks. Installation details and basic configuration of ROS environment, as well as installation and configuration of packages such as V-REP/ROS bridge, and the details of the Fanuc manipulator package are not in the concept of this chapter. A more detailed discussion can be found in [38].
V-REP is like a Swiss knife in robotic simulation community. Its first public release was in March 2010, and its latest version (V3.4.0 v1) was released on April 16, 2017. It possesses various relatively independent functions, features, or more elaborate APIs, that can be enabled or disabled as desired. Compared to gazebo, V-REP is very stable and easy to set up and running. For example, the vision sensors are reasonably well simulated and if the scene is not too complex, the run times of the simulations are generally good as well. If the project requires building a custom robot in the simulator (i.e., NOVABOT or Fanuc LR Mate 200iD manipulator), the setups for links, joints, and calculation modules such as inverse kinematics necessitates some practice, however, that is the case in any robot simulation software. Another big advantage is its true cross-platform, which means it can be run in Windows or Linux. By default, the V-REP distribution for Linux should be automatically ROS enabled based on ROS Indigo and Catkin. Each object/model in V-REP scene can be individually controlled via an embedded script, a plugin, a ROS node, a remote API client, or a custom solution. Controllers can be written in C/C++, Python, Java, Lua, Matlab, and Octaveor Urbi. The three main elements of V-REP simulator are scene object (i.e., joints, shape, sensors, path, etc.), calculation modules (i.e., inverse kinematics, collision detection, etc.), and control mechanism (i.e., scripts, plugin, sockets, etc.). In addition, V-REP inverse kinematics supports four different dynamic engines: The Bullet, ODE, Newton, and the Vortex Dynamics Engine. An overview of V-REP framework architecture is shown in Figure 4.
#### Figure 4.
Schematic diagram showing the architecture and the main elements of V-REP simulator.
## 3. Image processing, publishing and subscription
Quantification of fruits to estimate the time required for robotic harvesting is an intensive labor task that is either ignored in high density greenhouses or is carried out manually by the use of hand pickers. We proposed a low-cost robust sweet pepper fruit recognition and tracking system using stream RGB images. Main hardware and software components of the system included a laptop computer (Lenovo Intel(R) Core(TM) i5-6200 U CPU@2.30GHz, RAM 8.00GB, 64-bit OS Windows 10), a Logitech camera (C920 HD Pro USB 1080p), supplementary halogen lamps, Adafruit Ultimate GPS breakout module 66 channel w/10 Hz (NY, USA), and Arduino Uno Microcontroller board. The image processing algorithm was implemented in MATLAB and applies median filter and image segmentation method to remove color noise from the RGB images of pepper fruits taken in the lab experiments at different angles, positions, and light conditions disturbances (varying illumination and overlapping). Figure 5 shows: (A) original image, (B–D) red, green, and blue bands, (E) mask of only red object, (F) regions filled, (G) masked-red image, (H) extracting red component from the masked red image and applying median filter to filter out the noise, (I) convert the resulting grayscale image into a binary image and removing all pixels with a gray level value less than 3000, (J) masked image showing only red-detected object, (K) blob analysis, bounding the red objects in rectangular box and showing centroid. The image processing algorithm was validated using 78 images obtained from lab experiments and internet sources, with a recognition success rate of 94% and average recognition time of less than 2 s per image. Results of the image processing were sent from MATLAB to V-REP for simulation of visual servo control. For the actual experiment, color images of sweet pepper were acquired under natural daylight condition in different greenhouse environment in the presence of the halogen lamps. Each band of the RGB image was transferred as a 24-bit, 640 by 480 pixels matrix and was processed in real time by the custom built MATLAB application on the laptop computer. ROS was used for exchanging data between the simulated environment and the real workspace via its publish-and-subscribe architecture. Another 57 images were obtained for experimenting with different fruit and plant position scenarios. In addition, internet searched images of sweet pepper taken at different greenhouse environments were used to verify the reliability and to improve the accuracy of the algorithm. The image subscription and publishing was performed by having V-REP ROS enabled based on ROS Indigo and Catkin build. The general ROS functionality in V-REP is supported via a generic plugin “libv_repExtRos.so” or “libv_repExtRos.dylib.” It should be noted that plugins are loaded when V-REP is launched, and the ROS plugin will be successfully loaded and initialized only if “roscore” is running at that time. The plugin is open source and can be modified as much as needed in order to support a specific feature or to extend its functionality. Three of the main ROS package folders in the V-REP, (located in programming/ros_packages) are the “vrep_common,” “vrep_plugin,” and “vrep_joy” as shown in the left side of Figure 6.
#### Figure 5.
Demonstration of the steps in the robust image processing algorithm using edge detection with fuzzy-logic for identification and tracking of sweet pepper.
#### Figure 6.
Image publishing and subscribing in ROS, Left image: snapshot of the main ROS package folders in the V-REP, and right image: snapshot of the simulation environment in V-REP publishing an image to a ROS node.
The first package was used to generate the services and stream messages that were needed to implement the V-REP API functions, while the second is the actual plugin that was compiled to a “.so” file used by V-REP. The “vrep_joy” package enables interaction with a joystick. Having the services and stream messages in a separate package allows for other application to use them in order to communicate with V-REP via ROS in a convenient way. These packages were copied to the catkin_ws/src folder. The command “$roscd” was then used to check whether ROS is aware of these packages (e.g.,$ roscd vrep_plugin). After navigating to the catkin_ws, the command “$catkin_make” was used to build the packages and to generate the plugins. The created plugins were then copied to the V-REP installation folder to be used for image subscription and publishing. A new terminal was opened in Ubuntu for staring the ROS master using the command “$ roscore.” Another terminal was opened and was navigated to the V-REP installation folder to launch the V-REP simulator in Ubuntu by typing the command “$. /vrep.sh.” The entire procedure is summarized as these steps: (i) installing ROS Indigo on Ubuntu and setting up the workspace folder, (ii) copying “ros_packages” in V-REP into the “catkin_ws/src” folder, (iii) source “setup.bash” file, (iv) run “roscore” and “. /vrep.sh.” The two available nodes, “/rosout” and “/vrep” and the three topics “/rosout,” “/rosout_agg,” “/vrep/info” were checked using “$ rosnode list” and “$rostopic list” commands, respectively. In addition, the command “$ rosservice list” was used to advertise all the services. It should be noted that the only V-REP topic that was advertised was “info” publisher that started as soon as the plugin was launched. All other V-REP topics for publishing and subscribing images and sensors were individually enabled using Lua commands: “simExtROS_enablePublisher” and “simExtROS_enableSubscriber.” Moreover, to visualize the vision sensor stream images and data, the “$rosrun image_view image_view image:=/vrep/visionSensorData” and “$ rostopic echo/vrep/visionSensorData” were used, respectively. Snapshot of the simulation environment is shown in the right side of Figure 6.
## 4. Simulation scene and objects
Simulation scene in V-REP contains several elemental objects that are assembled in a tree-like hierarchy and operate in conjunction with each other to achieve an objective. In addition, V-REP has several calculation modules that can directly operate on one or several scene objects. Major scene objects and modules used in the simulation scene include (i) sensors, (ii) CAD models of the plant and robot manipulator, (iii) inverse kinematics, (iv) minimum distance calculation, (v) collision detection, (vi) path planning, and (vii) visual servo control. Other objects that were used as basic building blocks are: dummies, joints, shapes, graphs, paths, lights, and cameras (Figure 7). In this section, we provide description for the sensors and CAD models, and assign the next section to the calculation modules.
#### Figure 7.
Major scene objects used in the simulation.
### 4.1. Sensors
V-REP supports different vision sensors (orthographic and perspective type) and proximity sensors (Ray-type, pyramid-type, cylinder-type, disk-type, and cone- or randomized ray-type proximity sensors). It is possible to model almost any proximity sensor subtype, from ultrasonic to infrared. In addition it has built-in CAD models of several available commercial sensors such as Microsoft Kinekt, 2D and 3D laser scanners, blob detection camera, Hokuyo URG 04LX UG01, SICK S300, and TimM10 sensors. Other models can be built similarly based on combinations of different vision and proximity sensors. The V-REP model of each sensors used for this simulation is shown below its actual images in Figure 8 which include: Fish-eye RGB Axis 212 PTZ sensor (Figure 8A), Infrared Proximity Sensor Long Range-Sharp GP2Y0A02YK0F (Figure 8B), SICK TiM310 fast laser measurement scanner (Figure 8C), Fast Hokuyo URG-04LX-UG01 scanning Laser Rangefinder (Figure 8D), and Microsoft Kinect (Figure 8E).
### Figure 8.
Major sensors used in the experiments, first row image are the actual and the second row images are the simulated sensors.
The fish-eye RGB camera was added for fruit detection, tracking, and for visual servo control with a custom set of filters that were designed for the image processing algorithm in MATLAB and V-REP. Two color cameras were also added for tracking the scene and the position of the robot end-effector with respect to the fruit and plant in order to provide a wider view of the vision sensor. The V-REP model of the Microsoft Kinect sensor includes RGB and depth vision sensors, and was used in the scene to calculate the time needed for the laser signal to hit an object and bounce back to its source, creating in this way a three-dimensional representation of the object. Five different proximity sensors with different shapes were also experimented in the simulation, including: laser ray, pyramid, cylinder, disk, and randomized ray-type. The laser-scanner rangefinder was considered in the simulation to measure distance between an observer object (i.e., the robot gripper or the end-effector camera) and a target (i.e., fruit, plant, or obstacles). Typical range finders work based on time-of-flight (TOF) and frequency phase-shift technologies. The TOF method utilizes laser by sending a pulse in a narrow beam toward the object and measuring the time taken by the pulse to be reflected off and return to the sensor. The frequency-phase shift method measures the phase of multiple frequencies on reflection together with performing simultaneous math calculations to deliver the final measure. Rangefinders are available in V-REP in the form of vision-sensors and proximity sensors. For example, the Hokuyo URG-04LX-UG01 and the 3D laser scanner range finder use a ray-type laser proximity sensor. The V-REP model for Fast-3D laser scanner uses vision sensor with the filters as illustrated in Figure 9. It should be noted that vision-sensors-based rangefinders have high calculation speed but lower precision, while proximity-sensors-based rangefinders have higher prevision in calculating the geometric distance with relatively lower calculation speed.
### Figure 9.
Filter used by each of the two vision sensors in “Fast Hokuyo URG-04LX-UG01” V-REP model (top), and by the vision sensor in the “3D laser scanner Fast” V-REP model (bottom).
The CAD models of the sweet pepper plant, including stem system, leaves, and pepper fruits, as well as the single and multiple arms robot manipulators that were used in the simulation are shown in Figures 1013. The Fanuc LR Mate 200iD robot manipulator shown in Figure 11 is a compact six-axis robot with the approximate size and reach of a human arm. It combines best-in-class robot weight-load capacity with standard IP67 protection and outstanding FANUC quality. This makes the Fanuc LR Mate 200iD the best and most reliable mini robot for process automation in many industries. The maximum load capacity at wrist = 7 kg, repeatability = 0.02 mm, mechanical weight = 25 kg, and reach = 717 mm. The joints motion range and maximum speed are summarized in the operator manual [39]. As alternative innovative solutions, simple robots, including a platform with multiple linear actuators (Figure 12), and multiple SCARA robot arms (Figure 13) with multiple lower-cost cameras and grippers were also designed for simulation.
### Figure 10.
CAD models of the entire plant system: fruit, leaves, stem, calyx, and leaves.
### Figure 11.
Simulation scene setup with CAD models of the professional robot manipulator used in visual servo control experiment, left: Fanuc LR Mate 200iD, right: NOVABOT.
### Figure 12.
Simulation scene setup with CAD models of the multiple linear actuator robotic platform.
### Figure 13.
Simulation scene setup with CAD models of the multiple SCARA arm robotic platform.
## 5. Calculation modules
In order to setup the robot manipulator for different experiment, several calculation modules, including minimum distance calculation, collision detection, path planning, inverse kinematics, and different control mechanism were used in V-REP. Snapshot of the calculation modules is provided in Figure 14. V-REP control mechanism are divided into (i) local interfaces, including Embedded scripts, Plugins, Add-ons, and (ii) remote interfaces, including remote API clients, custom solutions, and ROS nodes, as shown in Figure 14A. It should be noted that different V-REP control mechanisms can be used simultaneously in one scene, or even work in conjunction with each other, which provides a multipurpose and accessible framework for the purpose of more complex robotic simulation. Scripting in V-REP is in the Lua language which is a fast scripting language designed to support procedural programming. Scripts in V-REP are the main control mechanism for a simulation. For the sake of this book chapter, we only provide brief illustration of the inverse kinematic task for the NOVABOT manipulator and the visual servo control.
#### Figure 14.
Demonstration of (A) five different control mechanisms in V-REP, (B) inverse kinematics chain, (C, D) minimum distance calculation from tip vision sensor and plant/fruit model, (E) collision detection between robot links and plant model, and (F) path planning for moving a harvested fruit to the bin.
### Figure 15.
Demonstration of the joint functionality and the inverse kinematics chain for the NOVABOT manipulator.
### 5.2. Visual servo control algorithm
A robot can be controlled in V-REP simulation through several ways such as child script, writing plugins, ROS nodes, external client applications that relies on the remote API, or writing an external application that communicates with V-REP plugin or script via pipes, sockets, or serial port. V-REP supports seven supported languages: C/C++, Python, Java, Matlab, Octave, Lua, and Urbi. In this research, we used MATLAB as the remote API because it provides a very convenient and easy way to write, modify and run image based visual servoing control codes. This also allows controlling a simulation or a model (e.g., a virtual robot) with the exact same code as the one that runs the real robot. The remote API functionality relies on the remote API plugin (on the server side), and the remote API code on the client side. Both programs/projects are open source (i.e., can be easily extended or translated for support of other languages) and can be found in the ‘programming’ directory of V-REP’s installation. Visual servo control scheme with eye-in-hand configuration, as shown in Figure 16, was implemented in MATLAB based on image moment method. For the case of the multiple linear actuators and the SCARA arms, we divided the camera view into multiple camera views to enhance the accuracy of the fruit detection algorithm and also to accelerate the image processing time (Figure 17). Details of the visual servo control algorithm are considered intellectual property of authors’ research group and are beyond the content of this chapter.
### Figure 16.
Visual servo control scheme with eye in hand configuration based on image moment method used with the Fanuc LR Mate 200iD and the NOVABOT manipulators.
### Figure 17.
Schematic diagram of an innovative approach for robotic harvesting based on multiple low-cost manipulators (e.g., multiple linear actuators or SCARA arms).
## 6. Results and discussions
Results provided a simulated environment for improvement of plant/fruit scanning and visual servoing task through easy testing and debugging of control algorithms with zero damage risk to the real robot and to the actual equipment. It also contributed to experimenting new ideas in robotic harvesting of sweet pepper, as well as testing different sensing instrumentation and control strategies on the currently used manipulators. Three groups of experiments, with separated V-REP scenes were designed for investigating different algorithms, robot manipulator, and sensors setup. They are summarized as experimenting with: (i) fruit detection and tracking algorithm using different camera views (Figures 18 and 19), (ii) manual and automated plant/fruit scanning in x-y, x-z, and y-z plane, and x-y-z space (Figures 20 and 21), (iii) fruit/plant scanning using Kinect, Hokuyo, fast 3D Laser, proximity 3D Laser scanner, and proximity Hokuyo URG04LXUG01 sensors (Figure 22), and (iv) visual servoing control law on single (Figure 23) and multiple (Figure 24) robot manipulator. Depending on the objectives of each scenario, sensors were placed in fixed spots, or on a moving link of the robot such as the end-effector. For example, the RGB vision sensor for fruit detection and tracking was used as eye-in-hand configuration with end-point closed-loop control. For the manual fruit/plant scan experiment with RGB sensors, the robot joints were controlled via sliders or by directly entering the desired joint angles in each label box as shown in Figure 20. This enabled sensing from the gripper from multiple viewpoints. In order to provide an interface with real workspace, two 2-axis analog Joysticks with push button were then used with Arduino Uno microcontroller to manually control angular positions for the joints. The automated fruit/plant scan experiments with RGB sensor were also carried out in different x, y, and z direction. The objective from this experiment was to simulate various camera pose and views for the best fruit attack and harvest. For Scanning in x-y plane, a 360° scan configuration of the fruit in the horizontal x-y plane is shown in Figure 20, with 30° increment snapshots of the simulated fruit. A similar scanning has been employed by [40]. For scanning in x-y-z space, two scan configurations in x-y-z space were used with snapshots of the resulting camera view shown in Figure 21. In this setup, the RGB sensor mounted on the robot tip is moved on the horizontal plane x-y to find the best view of the fruit. Moreover, the manipulator is “twisted” to provide different viewpoints for the end-effector camera.
#### Figure 18.
Results of the image processing algorithm for fruit localization using wide camera view (left) and divided camera view (right).
#### Figure 19.
Result of the image processing algorithm for quantification and tracking of sweet pepper in different fruit-and-plant scenario.
#### Figure 20.
Two dimensional scanning experiment (x-y, x-z, and y-z planes) for finding the maximum fruit visibility. Camera was set to move at 30 degrees increments around the fruit and plant model.
#### Figure 21.
Three dimensional scanning experiments (x-y-z space) for finding the maximum fruit visibility. Camera was set to rotate around the fruit and plant until it finds the best angle of attack.
#### Figure 22.
Simulation of scanning experiment with Laser scanners and depth sensors.
#### Figure 23.
Simulation of visual servo control experiment with the eye-in-hand configuration and PID Control law on joint angles with feedbacks from image moments. Stability was achieved in 2.5 s without overshoot and oscillations.
#### Figure 24.
Simulation of robotic harvesting with arrays of linear actuators.
The “3D Laser Scanner Fast” sensor model in V-REP is based on vision-sensor with a perspective angle equal to 45°, a resolution of 512 by 512 and minimum and maximum 0.05 and 5 m distance of operation. Snapshot of the experiment with this sensor is shown in Figure 22. The “Fast Hokuyo URG-04LX-UG01” model in V-REP also works in perspective mode with an operability angle equal to 120°, and a resolution that was set at 512 by 1 which means it scans along a line shown on the floating view. It has a minimum and maximum distance of operability, respectively, equal to 0.04 and 5. The image processing in this case is similar to the 3D laser sensor except that the intensity map scale component is omitted. This sensor in fact does not come with any floating view by default. Two floating views were added for the two vision sensors of the “Fast Hokuyo URG-04LX-UG01” model. The black line inside the floating view of each sensor indicates presence of object (i.e., fruit, leaf, robot, and plant). First row images in Figure 22 are snapshot of the scene with Kinect depth sensor in action for fruit/plant scan, and the bottom row images are, respectively, from left to right are: snapshot of the scene with vision sensors, showing the “3D Laser scanner Fast,” the “Fast Hokuyo URG-04LX-UG01,” snapshot of the scene with proximity sensors showing the “3D-Laser scanner,” and the “Hokuyo URG-04LX-UG01” scanning scene.
For the visual servo control task, a robot end-mounted camera was used to position the robot arm in a plane orthogonal to the axis, such that the fruit to be harvested is centered in the camera’s field of view. The system had no trajectory generator, instead a feedback loop closed visually, was used to control the robots arm position. The sensor and robot was programmed for visual servoing task in such a way that the end-effector tracks the largest detected fruit until maximum possible view of that fruit is provided. Two different control approaches was designed and tested, one based on joint velocity control and the other based on joint position control. In both design a PID control law was implemented to minimize the offset error between image position of a detected fruit and center of the camera frame. Results showed that the robot could adjust itself in such a way that its tip RGB sensor shows maximum possible view of the largest detected fruit and become stable in a short time. It should be noted that both control algorithms were designed and tuned based on the experiments and statistical results from fruit/plant scan. Video demonstration of the entire experiments can be accessed from the links listed in Table 2.
Simulation of NOVABOT innovative manipulatorhttps://youtu.be/R38IoVcOVt0
Simulation of multiple SCARA armshttps://youtu.be/TLLW3S-55ls
Simulation of multiple linear actuatorshttps://youtu.be/iFu7FAxLvmg
Robotic Harvesting with Fanuc LR Mate 200iDhttps://youtu.be/BwRBUeB812s
Simulation of Environment mapping and scanninghttps://youtu.be/XD3J7b0cDGM
Detailed demonstration of fruit and plant scanhttps://youtu.be/6EOy1NesvQg
Detailed demonstration of visual servo controlhttps://youtu.be/VupoirQOL0Y
Testing Visual Servo Control Algorithmhttps://youtu.be/VupoirQOL0Y
Environment Setup: Ubuntu, V-REP, ROShttps://youtu.be/tKagjNQ9FW4
Real-time, robust and rapid red-pepper fruit detectionhttps://youtu.be/rFV6Y5ivLF8
#### Table 2.
As the final discussion, we would like to highlight that agricultural robotic is a part of the big picture in the future production of vegetable and crops, i.e., growing plants in space. An example includes space research for development of Mars greenhouses to produce vegetables during a mission to Mars or at Antarctica. The trend in food production is toward urban farming techniques, compact Agri-cubes, and automated systems with minimum human interface. The idea is that even people with limited experience/knowledge in vegetable cultivation can operate these units. While this integration might seem too ambitious, it can serve as a prophetic awareness for a perceptive outlook in the farming system. For example, the conventional arrangements of citrus groves, orchards, and the trees shapes in Florida had to be reconsidered for the mechanical harvesting machines to operate successfully. It is likely that the greenhouse planting systems for sweet pepper will also be re-shaped to match with a customized robotic platform. Two of the key challenges to be solved during the design of robotic harvesting framework are addressed by [40] as (i) detection of a target location of the fruit, and (ii) moving the end-effector toward that location with precision for harvesting task. We argue that these two challenges have not been stated accurately. First of all, it is not always necessary to detect the target location of the fruit, especially in the case of a mass harvesting platform with shaking grippers. Second, moving the end-effector toward the target fruit is not a scientifically sound statement, e.g., considering the strategy case in which the plant system is moved toward a fixed end-effector. To avoid this divergence, the task should have been stated as minimizing the error between the location of the end-effector and the target fruit. In fact, a promising solution to the robotic harvesting is not through a single robot manipulator. We provided a quick review of the previous works, and used simulation approach to reveal that single arm robots for harvesting are still far beyond realization, and have failed mainly due to the sensing and moving action in high vegetation density. In this approach, even if the fruit localization is accurate, and the robot control calculates an optimum trajectory to reach the fruit without receiving additional sensing feedback from the camera, the moment it enters into the dense plant canopy it disrupts the exact location of the target fruit.
## 7. Conclusion
Research and development for the use of robotics in agriculture that can work successively have grown significantly in the past decade; however, to this date, a commercial robotic harvesting is still unavailable for fresh fruit market. With the decrease of greenhouse workforce and the increase of production cost, research areas on robotic harvesting have received more and more attention in recent years. For the success of robotic harvesting, the identification of mature fruit and obstacle is the priority task. This chapter reported on a simulation and control platform for designing, testing, and calibration of visual servoing tasks in robotic harvesting of sweet-pepper. Creation of a virtual environment was carried out as a response to the improvement of fruit detection rate. We provided a documented guideline for a reliable, cheap and safe experiment platform with a faster approach for development, testing, and validating control strategies and algorithms to be used with different robot candidates and gripper mechanism in automated harvesting of fruiting vegetables. Results of the image processing confirmed that the presented approach can quantify and track mature red sweet pepper fruits from its surrounding obstacles in the real-time. It can be concluded that development of an affordable and efficient harvesting robot requires collaboration in areas of horticultural engineering, machine vision, sensing, robotics, control, intelligent systems, software architecture, system integration, and greenhouse crop management. In addition, practicing other cultivation systems in the greenhouse, such as single row, might be necessary for overcoming the problems of fruit visibility and accessibility. It can also be concluded that human-robot collaboration might be necessary to solve the challenges in robotic harvesting that cannot yet be automated. In a collaborative harvesting with human-robot interface, any fruit that is missed by the robot vision is identified by the human on the touch screen, or the entire robot actions are controlled manually in a virtual environment. Nevertheless, robotic harvesting must be economically viable which means it must sense fast, calculate fast, and move fast to pick a large number of fruits every hour that are bruise free.
## References
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10 - Tao Y, Zhou J. Automatic apple recognition based on the fusion of color and 3D feature for robotic fruit picking. Computers and Electronics in Agriculture. 2017;142(Part A):388-396
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13 - Bac Y, Henten CW, Hemming EJ, Edan J. Harvesting robots for high-value crops: State-of-the-art review and challenges ahead. Journal of Field Robotics. 2014;31(6):888-911
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15 - Van Henten J, Van Tuijl EJ, Hoogakker BAJ, Van Der Weerd GJ, Hemming MJ, Kornet J, Bontsema JG. An autonomous robot for de-leafing cucumber plants grown in a high-wire cultivation system. Biosystems Engineering. 2006;94(3):317-323
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23 - Tang Y, Zhang X, Liu T, Xiao L, Chen D. A new robot system for harvesting cucumber. In: American Society of Agricultural and Biological Engineers Annual International Meeting; 2009. pp. 3873-3885
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28 - Mehta SS, Burks TF. Vision-based control of robotic manipulator for citrus harvesting. Computers and Electronics in Agriculture. 2014;102:146-158
29 - Senthilnath J, Dokania A, Kandukuri M, Ramesh KN, Anand G, Omkar SN. Detection of tomatoes using spectral-spatial methods in remotely sensed RGB images captured by UAV. Biosystems Engineering. 2016;146:16-32
30 - De-An Z, Jidong L, Wei J, Ying Z, Yu C. Design and control of an apple harvesting robot. Biosystems Engineering. 2011;110(2):112-122
31 - Li Z, Li P, Yang H, Wang Y. Stability tests of two-finger tomato grasping for harvesting robots. Biosystems Engineering. 2013;116(2):163-170
32 - Bac CW, Roorda T, Reshef R, Berman S, Hemming J, van Henten EJ. Analysis of a motion planning problem for sweet-pepper harvesting in a dense obstacle environment. Biosystems Engineering. 2016;146:85-97
33 - Bloch V, Degani A, Bechar A. A methodology of orchard architecture design for an optimal harvesting robot. Biosystems Engineering. 2018;166:126-137
34 - Hellström O, Ringdahl T. A software framework for agricultural and forestry robots, Industrial Robot. Industrial Robot: An International Journal. 2013;40(1):20-26
35 - Amatya S, Karkee M, Gongal A, Zhang Q, Whiting MD. Detection of cherry tree branches with full foliage in planar architecture for automated sweet-cherry harvesting. Biosystems Engineering. 2015;146:3-15
36 - Barnea E, Mairon R, Ben-Shahar O. Colour-agnostic shape-based 3D fruit detection for crop harvesting robots. Biosystems Engineering. 2016;146:57-70
37 - Quigley M, Conley K, Gerkey B, Faust J, Foote T, Leibs J, Berger E, Wheeler R, Mg A. ROS: an open-source Robot Operating System. ICRA. 2009;3:5
38 - Bouchier P. Embedded ROS [ROS Topics]. IEEE Robotics and Automation Magazine. 2013;20(2):17-19
39 - FANUC Robot LR Mate 200ic mechanical unit operator’s manual. MAROTLR2006071E REV. G, B-82584EN/06. Rochester Hills, Michigan: FANUC Robotics America, Inc.; 48309-3253. Available at: http://www.msamc.org/aimss/documentation/pdf/manuals/lr_mate_manuals/LR%20Mate%20200iC%20Operators%20Manual.pdf [Last access: February 10th, 20]
40 - Barth R, Hemming J, van Henten EJ. Design of an eye-in-hand sensing and servo control framework for harvesting robotics in dense vegetation. Biosystems Engineering. 2016;146:71-84
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2018-03-23 07:33:04
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https://cs.stackexchange.com/questions/56074/finding-the-language-generated-for-cfg
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# Finding the language generated for CFG
What language generated by the following context-free grammar 1) S------> SaS | b i already know the answer to question one but to prove it would is be something like this: S -----> SaaS -----> baab which can also be express as the language b(ab)* or (ba)*b
Another question is what does | b mean does it mean substitute into S afterwards?
2) S ----> AA
A----> aA | Aa | b
So for this question how can i start this? Would it be like S----> AA ----> aAaA ----> aAaAaA -----> abababa ? SO the language would be (ab)*?
• One question per question, please. The answer to many of these is covered by standard textbooks and by our reference materials. For instance, the ansewr to your first question (how to prove what the language is) is covered by cs.stackexchange.com/q/11315/755. Note that you can use LaTeX here to typeset mathematics in a more readable way. See here for a short introduction. – D.W. Apr 17 '16 at 19:43
• Your question is a very basic one. Let me direct you towards our reference questions which cover some fundamentals you seem to be missing in detail. Please work through the related questions listed there, try to solve your problem again and edit to include your attempts along with the specific problems you encountered. Good luck! – Raphael Apr 17 '16 at 20:26
• For the language in (2), note that $A$ can only derive strings with a single $A$, so $S$ can only derive strings of two $A$s, which will eventually yield strings with exactly two $b$s, so it can't be the language denoted by $(ab)^*$. – Rick Decker Apr 17 '16 at 23:42
• @Rick Decker would the answer in #2 be (a+b)* ? since it can start with a or b and end with a or b which is almost similar to (ab)* – Darkflame May 1 '16 at 18:44
• The language denoted by $(a+b)^*$ is not the same as the one denoted by $(ab)^*$. The former is all strings over $\{a,b\}$ and the latter is $\{\epsilon,ab,abab,ababab,\dotsc\}$. As I mentioned above, the language generated by your grammar consists of all strings with exactly two $b$s. – Rick Decker May 2 '16 at 12:47
The vertical bar indicates an alternative: $S\to SaS \mid b$ means there are two production rules in the grammar, $S\to SaS$ and $S\to b$. Since $b$ contains no variables it is like a final rule applied to a particular branch of the derivation-tree. But that does not mean the rule $S\to b$ has to be the postponed and applied to all occurrences of $S$ in the end.
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2020-06-02 12:29:30
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https://www.physicsforums.com/threads/unseparable-o-d-e.86516/
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Unseparable O.D.E.
1. Aug 27, 2005
asdf1
another O.D.E. question:
xyy= 2y^2 +4x^2
also unseparable...
2. Aug 27, 2005
Galileo
Im a bit rusty on D.E.'s, but you can write it as:
$$y'=2(y/x)+4(x/y)$$
IIRC, whenever you have y'=F(y/x), the equation also called homogeneous (different from the homogeneous linear ODE definition), this one is invariant under 'zoom'.
If you make the subsitution z=y/x, the equation becomes:
$$z+xz'=2z+4/z$$
or
$$xz'=z+4/z$$
which is separable.
Last edited: Aug 27, 2005
3. Aug 28, 2005
GCT
it's separable, use v=y/x
4. Aug 29, 2005
asdf1
why'd you think of v=y/x?
5. Aug 29, 2005
TD
Galileo explained. It's because the DE is homogeneous (not with the 0 constant, but with the sum of all exponents of all terms is constant (here 2). You can then divide so the DE becomes a function of (y/x). Susbtitution will make it separable.
6. Aug 29, 2005
asdf1
so the first step when you see an O.D.E you have to check if it's separable
=> if not, then check if it's homogeneous, and use that substitution...
thanks! :)
7. Aug 29, 2005
TD
Checking if it's separable should indeed always be the first you do and when it's not, try to see if you can make it separable with a substitution.
The substitution used here a very common one.
8. Aug 30, 2005
asdf1
what are the most common substitutions used?
9. Aug 31, 2005
Benny
Well y = v(x)x is a common one and if I recall correctly, for second order isobaric equations you can make the substitution y = vx^m if there is a single value of m which makes the equation dimensionally consistent (you assign weights to y, dy, x and dx).
Last edited: Sep 1, 2005
10. Aug 31, 2005
asdf1
ok~ thanks!!! :)
11. Sep 3, 2005
asdf1
hmmm... here's my work~
(i forgot to include that the original condition is y(2)=4, sorry about that!)
y= 2y/x+4x/y
suppose y=vx
=>v=y/x
so y=v+vx
=> v+xv=2v +4/v
=> vx=v+4/v
=> vx=(v^2 +4)/v
=> v/(v^2 +4)dv=dx
=> 1/2[dv^2/(v^2 +4)] =dx
=> 1/2ln[absolute value (v^2 +4)] =x+ c
=> ln[absolute value (y^2/(x^2) +4)]=2x+c
=> y^2/(x^2) +4 =e^(2x +c`)
=> y^2 +4x^2 =c(x^2)*[e^(2x)]
=> y^2=c(x^2)*[e^(2x)]-4x^2
now my problem is that when you take the square root on both sides of the equations, the right side should have a positive and negative root, but the correct answer should only have a positive square root~
why?
btw, what is meant by the different homogeneous definitions and "zoom"?
12. Sep 3, 2005
Galileo
Whoopsie, you've lost an x.
The actual answer also needs a square root since you'll get a relation between y^2 and x. Now if this equation is true both the positive and negative square roots must be valid solutions to the ODE, but a little bit of thought shows that no more than one can fit the same boundary condition (y(2)=4)
A linear DE of, say, second order:
$$y''+A(x)y'+B(x)y=r(x)$$
is called homogeneous if r(x)=0. Your ODE is a also called homogeneous, but it's in a different context, so it naturally hasn't the same meaning.
Since the ODE can be written in terms of y/x. If we make the change of variables $\hat y=ay$ and $\hat x=ax$ for some positive a. So that we're scaling the y and x-axes by the same amount (this is what I meant by zoom), the ODE looks the same, since $\hat y/\hat x=y/x$.
13. Sep 4, 2005
asdf1
thank you very much!!! :)
i'll try and be more careful on my calculations~
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2017-10-21 22:26:18
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https://math.stackexchange.com/questions/3069744/the-motivation-of-continuous-random-variable-mutual-information-by-kth-neares
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# The motivation of continuous random variable mutual information by $k$'th nearest neighbour
I am reading this paper Kraskov et al, 2004, Estimating Mutual Information on estimating the mutual information of two continuous random variables based on entropy estimates from $$k$$-nearest neighbour distances. I do not quite understand the motivation is using $$k$$-nearest neighbour distances. Could someone elucidate it?
This paper is also referred to in this answer.
• Is the question that you want the intuition behind it? – user3658307 Jan 12 at 2:38
• @user3658307: Yes, the intuitive motivation. – Hans Jan 12 at 2:57
Here's the intuition from my perspective.
First at a high level, the mutual information (MI) is measuring the (expected, logarithmic) distance between the joint distribution $$p(x,y)$$ and the product of the marginals $$p(x)p(y)$$. This means you need some kind of way to estimate densities. The obvious way is to make a histogram, and count the (normalized) number of points falling into each bin.
However, keep in mind that we are ultimately interested in the probability density, which is directly related to the sample density. But, just intuitively, samples are dense in a particular location if there are many of them close together; i.e., the distances to their nearest neighbours are small. In other words, areas of high density have densely packed samples, meaning the distance between nearby points is small. Conversely, areas of low density have few samples, and so the distance to their closest neighbours will be large.
Hence, the paper tries to
estimate MI from k-nearest neighbour statistics
So basically we can just think of using neighbour distances as another proxy for sample density, instead of counting samples in bins. It sort of reminds me of kernel density estimation.
More concretely, for this paper, however, they are thinking more about a different perspective on the mutual information, namely as the difference between the independent (marginal) entropies and the joint entropy. They then use estimators for the entropies, based on the neighbour statistics. Another intuition, I think, is that counting samples based on local neighbourhood distances (in the joint space) makes the algorithm adaptive (as opposed to choosing a single bin size over the whole sample space).
• What is "it" referring to in "It sort of reminds me of kernel density estimation", the $k$-nearest neighbour or counting samples in bins? To me, the kernel density estimation is akin to the histogram only that this histogram is overlapping and "soft". What puzzles is then which $k$ one should choose? Should we sum or average over all $k$'s or leave it as a hyperparameter to be determined by, say, cross-validation? Also, which is better, $k$-nearest neighbour or kenel density estimation? – Hans Jan 12 at 7:40
• By the way, you may be interested in and have an opinion on my stats.stackexchange.com question stats.stackexchange.com/q/386101/44368. – Hans Jan 12 at 7:45
• @Hans $k$ is chosen as a fixed value, I believe. Some discussion of the tradeoff of higher vs lower $k$ is talked about in the paper (e.g., computational cost, relation to the scale of the data). You have to choose $k$ based on these considerations. KDE is indeed like a soft histogram. "It" was just referring to the nearest neighbour method. KDE also has several (difficult) parameters to choose: e.g., which kernel? which bandwidth? I am not an expert in this area, but I suspect that both methods can be accurate in practice, but it depends on the chosen parameters and the data :) – user3658307 Jan 12 at 14:52
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2019-01-21 09:08:50
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https://socratic.org/questions/sales-ling-can-spend-no-more-than-120-at-the-summer-sale-of-a-department-store-s
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# SALES Ling can spend no more than $120 at the summer sale of a department store. She wants to buy shirts on sale for$15 each. How do you write and solve an inequality to determine the number of shirts she can buy?
Apr 29, 2018
The inequality is $15 x \le 120$, and she can buy at most $8$ shirts.
#### Explanation:
Let's split this word problem up.
"no more" refers to the number or less than that, or $\le$.
So whatever Ling buys has to be $\le 120$.
She wants to buy an UNKNOWN number of shirts for $15 each. So we set that UNKNOWN value to $x$, and form an inequality: $15 x \le 120$To solve for $x$, we divide both sides by $\textcolor{red}{15}$: $\frac{15 x}{\textcolor{red}{15}} \le \frac{120}{\textcolor{red}{15}}$Therefore, $x \le 8$She can buy at most $8\$ shirts.
Hope this helps!
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2019-10-18 02:37:33
|
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|
http://eprints.pascal-network.org/archive/00007952/
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Hilbert Space Embeddings and Metrics on Probability Measures
Bharath Sriperumbudur, Arthur Gretton, Kenji Fukumizu, Bernhard Schölkopf and Gert Lanckriet
JMLR Volume 11, pp. 1517-1561, 2010.
## Abstract
A Hilbert space embedding for probability measures has recently been proposed, with applications including dimensionality reduction, homogeneity testing, and independence testing. This embedding represents any probability measure as a mean element in a reproducing kernel Hilbert space (RKHS). A pseudometric on the space of probability measures can be defined as the distance between distribution embeddings: we denote this as $\gamma_k$, indexed by the kernel function $k$ that defines the inner product in the RKHS. \par We present three theoretical properties of $\gamma_k$. First, we consider the question of determining the conditions on the kernel $k$ for which $\gamma_k$ is a metric: such $k$ are denoted {\em characteristic kernels}. Unlike pseudometrics, a metric is zero only when two distributions coincide, thus ensuring the RKHS embedding maps all distributions uniquely (i.e., the embedding is injective). While previously published conditions may apply only in restricted circumstances (e.g. on compact domains), and are difficult to check, our conditions are straightforward and intuitive: \emph{integrally strictly positive definite kernels} are characteristic. Alternatively, if a bounded continuous kernel is translation-invariant on $\bb{R}^d$, then it is characteristic if and only if the support of its Fourier transform is the entire $\bb{R}^d$. Second, we show that the distance between distributions under $\gamma_k$ results from an interplay between the properties of the kernel and the distributions, by demonstrating that distributions are close in the embedding space when their differences occur at higher frequencies. Third, to understand the nature of the topology induced by $\gamma_k$, we relate $\gamma_k$ to other popular metrics on probability measures, and present conditions on the kernel $k$ under which $\gamma_k$ metrizes the weak topology.
EPrint Type: Article Project Keyword UNSPECIFIED Theory & Algorithms 7952 Arthur Gretton 17 March 2011
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2014-11-23 09:52:30
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https://blender.stackexchange.com/questions/106074/changing-image-texture-as-height-increases-and-using-colorramp-with-image-textur
|
# Changing image texture as height increases and using colorRamp with image texture
I am trying to make a landscape scene. I want the landscape to have 4 different images as textures. The images should change as the height increases. I was able to do this with 2 images but when I increase the images it goes wrong. Also I found examples on stack which were using color ramp. Also when I use a mapping node for my images, the result is having only one Image. I use cycles engine.
I want something like this but instead of colors i want to use images
Use different ramps, each one with a different increasing range, to map the information from the $Z$ axis.
|
2019-12-07 11:55:29
|
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https://research.chalmers.se/en/publication/227957
|
# Segre numbers, a generalized King formula, and local intersections Journal article, 2015
Let $\mathcal{J}$ be an ideal sheaf on a reduced analytic space $X$ with zero set $Z$. We show that the Lelong numbers of the restrictions to $Z$ of certain generalized Monge– Ampère products $(dd^c \log |f|^2)^k$, where $f$ is a tuple of generators of $\mathcal{J}$, coincide with the so-called Segre numbers of $\mathcal{J}$, introduced independently by Tworzewski, Achilles–Manaresi, and Gaffney–Gassler. More generally we show that these currents satisfy a generalization of the classical King formula that takes into account fixed and moving components of Vogel cycles associated with $\mathcal{J}$. A basic tool is a new calculus for products of positive currents of Bochner–Martinelli type. We also discuss connections to intersection theory.
## Author
University of Gothenburg
Chalmers, Mathematical Sciences, Mathematics
#### Håkan Samuelsson Kalm
University of Gothenburg
Chalmers, Mathematical Sciences, Mathematics
#### Elizabeth Wulcan
Chalmers, Mathematical Sciences, Mathematics
University of Gothenburg
#### Alain Yger
University of Bordeaux
#### Journal für die Reine und Angewandte Mathematik
0075-4102 (ISSN)
Vol. 728 728 105-136
#### Subject Categories
Mathematics
Geometry
Mathematical Analysis
#### DOI
10.1515/crelle-2014-0109
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2019-08-22 13:41:16
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https://blender.stackexchange.com/questions/5225/accessing-plane-track-using-python-api
|
# Accessing plane track using Python API
I'm new to Blender and the Python language and I'm trying to extract the coordinates of the corners of a plane track from all the frames of a clip. I'm having trouble even accessing(?) the plane track object/variable. I hover over the text field containing the name of the plane track and it says
...
bpy.data.movieclips["1340wbdc0000.jpeg"].tracking.plane_tracks["pt"].name
I type that line into the console, but then I get the error
Traceback (most recent call last):
File "<blender_console>", line 1, in <module>
However, if I enter
D.movieclips["1340wbdc0000.jpeg"].tracking.plane_tracks.active
The console "responds" with
bpy.data.movieclips['1340wbdc0000.jpeg'].tracking.plane_tracks["pt"]
This is really making me very confused and frustrated. What am I doing wrong? Could this be a bug since plane tracking is a new feature?
Works for me in 2.69:
>>> bpy.data.movieclips[0].tracking.plane_tracks.active.markers[0].corners[0][:]
#(0.12633077800273895, 0.4383348524570465)
plane_tracks.active gives me the correct name, even if I rename it in the right sidebar of the movieclip editor. You could try plane_tracks["Plane Track"] or plane_tracks[0] instead.
Do should deselect all in the movieclip editor and re-select the tracking plane and see if plane_track.active changes to the correct name.
clip = bpy.data.movieclips["1100wbdc0000.jpeg"]
print("\nMovie Clip %s" % clip.name)
for i, ob in enumerate(clip.tracking.objects):
print("\n\tTracking Object %s" % ob.name)
for j, pt in enumerate(ob.plane_tracks):
print("\t\tPlane Track %i" % j)
for k, marker in enumerate(pt.markers):
print("\t\t\tMarker %i" % k)
for l, corner in enumerate(marker.corners):
print("\t\t\t\tCorner %i = %f / %f" % (l, corner[0], corner[1]))
• Yes, plane_tracks.active does give me the correct name. However if no plane track is selected, using plane_tracks["Plane Track"] or plane_tracks[0] gives me an error. I was hoping to write a script to automate the process since I have around 60 plane tracks to extract. – user1726 Dec 2 '13 at 23:47
• Of course it gives an error if there's no active plane track. If you need to access all plane tracks, why not loop over all of them? There's no need to make them / one active. for pt in Clip.tracking.plane_tracks: pt.markers... – CoDEmanX Dec 3 '13 at 20:18
• That's my problem. I can't loop over all of them. I tried the following code: import bpy i=0 clip=bpy.data.movieclips["1100wbdc0000.jpeg"] print(clip.tracking.plane_tracks.active.name) for track in clip.tracking.plane_tracks: print(i) i=i+1 – user1726 Dec 4 '13 at 10:53
• If a plane track is selected, Plane Track 0 is printed on the console. However, if no plane track is selected, it returns AttributeError: 'NoneType' object has no attribute 'name' – user1726 Dec 4 '13 at 11:01
• There's no point in accessing plane_tracks.active if you wanna loop for all Plane Tracks, their markers and their Corners. See my answer above, added example code. – CoDEmanX Dec 4 '13 at 15:49
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2019-07-19 09:55:21
|
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|
https://www.transtutors.com/questions/e11-11-depreciation-change-in-estimate-tom-brady-co-86563.htm
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# E11-11 Depreciation—Change in Estimate Tom Brady Co
1 answer below »
Week Four (Week 4)
Chapter 11 and Chapter 12
E11-11 (Depreciation—Change in Estimate) Machinery purchased for $60,000 by Tom Brady Co. in 2003 was originally estimated to have a life of 8 years with a salvage value of$4,000 at the end of that time. Depreciation has been entered for 5 years on this basis. In 2008, it is determined that the total estimated life should be 10 years with a salvage value of \$4,500 at the end of that time. Assume straight-line depreciation.
Instructions
(a) Prepare the entry to correct the prior years’ depreciation, if necessary.
(b) Prepare the entry to record depreciation for 2008.
## 1 Approved Answer
5 Ratings, (9 Votes)
a
There is no correction needed
Cost in whole 60,000 Salvage value 4,000 Depreciable...
## Recent Questions in Financial Accounting
Submit Your Questions Here!
Copy and paste your question here...
Attach Files
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2018-10-19 02:18:49
|
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https://samyzaf.com/ML/opens/opens.html
|
# VLSI Circuit Open Defects Detection¶
Open defects in VLSI integrated circuit production processes are known to be a common issue. They are usually sought after in the Post Silicon manufacturing stages or when defective chips are returned by customers. Soon after the first wafers come out from the fabrication factories, special test teams are hunting for circuit opens with the aid of highly expensive tester and prober equipments which include high resolution microscopes and cameras that can quickly scan large areas of the chip and produce high quality photos of suspected areas of the chip.
Through the years, VLSI research and development centers introduced advanced methods to process and analyze these photos in order to find defects such as opens or shorts. No doubt that after entering the age of deep learning, there ought to be attempts to attack this problem from the deep learning point of view, which has already proved to be very successful in computer vision.
As this tutorial is intended for a non-expert audience whose main interest is focusing on deep learning with neural networks, we will use a simplistic view of the subject so that we are not distracted from our main interest on showing how deep learning can be applied in the VLSI domain as well as in other domains such as computer vision, natural language processing and so on.
## The circuit opens dataset¶
We have generated a simplistic dataset of small defective circuits (due to an open). Each circuit is drawn on a small 48x48 pixels black/white image, and is supposed to contain at least one open defect. The dataset is divided to 5 HDF5 files, each containing 100,000 images. So we have a total of 500,000 images of defective circuits. We hope they are sufficient for training a neural network to do the job (but we must keep a small portion for validation too).
For each circuit i, there are 3 entries associated in the dataset:
* img_i - A Numpy matrix for the image itself
* open_i - The open defect rectangular area
* center_i - an (x,y) coordinates of the defect center
## Agenda¶
Our aim in this course unit is to suggest a project proposal for applying deep learning methods to VLSI defects detection, based on our simple dataset as a simple example on how to do it. As a student, your goal is to define a deep convolutional neural network that accept a circuit image as input and finds the point in which the circuit is open.
In [1]:
# These are css/html styles for good looking ipython notebooks
from IPython.core.display import HTML
HTML('<style>{}</style>'.format(css))
Out[1]:
## Feature Information¶
We suggest that a circuit open defect should be coded by a pair of integers which is the coordinate of its center point. This is only a simple suggestion. You may offer a better one if you like. So the features list is the list of all points (i,j) in a 48x48 pixels image space of our circuit.
In [2]:
features = [(i,j) for i in range(48) for j in range(48)]
Here is a simple diagram which explains this idea
Note the Numpy coordinate convention: the first coordinate i designates the row number, the second coordinate j designates the column number. This is also the standard convention in linear algebra matrix notation.
So in total we will have 48x48 features (i.e., 2304 features), which might be too challenging for the training phase. The neural network output layer should have 2304 output neurons, which is a bit discouraging. You may want to refine the list of features by a simple list of regions in which the defect occurs. Once we know the region, it will be easy to locate the defect. Here is an example in which our 48x48 image is divided to 64 6x6 regions
With this scheme, our neural network will have only 64 output neurons, which is not too bad. If you choose to work with this scheme, you will have to convert each defect point in our dataset to the corresponding region number in the training phase. It shouldn't be hard to find a simple formula for doing it. But again, feel free to explore other schemes (you may be forced to do so in case that your training dataset is insufficient for hitting precise locations of defects, so you might need to resort to coarser subdivisions)
In [3]:
# If you stick to the full 48x48 features scheme, then you will have 2304 classes
# Here is a simple way you can map pairs (i,j) to integers
classes = [48*i+j for i in range(48) for j in range(48)]
# This is a very long list, so we print the first 100 elements only:
print(classes[0:100])
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99]
## Data Exploration¶
We first show how to parse an HDF5 in order to extract the image and other circuit data.
In [4]:
from __future__ import print_function # we want to enable Python 3
import h5py
import numpy as np
# This is how you open an HDF5 file
f = h5py.File('opens1.h5', 'r')
# We will collect 2000 images (from th 80,000 images in this file) in one long Python list
images = []
# We will collect the first 2000 boxes
boxes = []
# And finally we will also collect the first 2000 defect centers
centers = []
# Our HDF5 file has a special key for the number of images which we're not going to use here
# but you will need it for the full project
# num_images = f.get('num_images').value
for i in range(2000):
box_key = 'open_' + str(i)
img_key = 'img_' + str(i)
center_key = 'center_' + str(i)
# This is the shape of the open defect
box = np.array(f.get(box_key))
boxes.append(box)
# This is image i
img = np.array(f.get(img_key))
images.append(img)
# This is the defect center data
c = np.array(f.get(center_key))
centers.append(c)
# Do not forget to close the HDF5 file!
f.close()
In [5]:
import matplotlib.pyplot as plt
from matplotlib import rcParams
%matplotlib inline
def draw_circ(i):
print("Image id:", i)
#print("Features:", hands[i])
#c = box[i]
img = images[i]
x, y = centers[i]
box = boxes[i]
print("Box: %s" % (box,))
print("Center: (%d,%d)" % (x,y))
plt.title("circ%d: open=(%d,%d)" % (i,x,y), fontsize=18, fontweight='bold', y=1.02)
ticks=[0,8,16,24,32,40, 48]
plt.xticks(ticks, fontsize=12)
plt.yticks(ticks, fontsize=12)
plt.imshow(img, cmap='gray', interpolation='none')
plt.plot(y, x, 'or')
plt.grid()
Let's draw a few samples of circuits.
In [6]:
draw_circ(35)
Image id: 35
Box: [25 27 17 19]
Center: (26,18)
In [7]:
draw_circ(476)
Image id: 476
Box: [20 22 27 29]
Center: (21,28)
In some cases we need to view a larger group of samples at the same time. For example if we have a large group of false predictions, it is useful to look at a dozen or two of them in order to get a feeling about the kind of failure. The following function we draw a grid of mxn of circuits, m rows, n columns:
In [8]:
rcParams['figure.figsize'] = 9,9
def draw_group(imgs, cntrs, rows=4, cols=4):
for i in range(0, rows*cols):
ax = plt.subplot(rows, cols, i+1)
img = imgs[i]
x, y = cntrs[i]
plt.title("circ%d: open=(%d,%d)" % (i,x,y), fontsize=8, fontweight='bold', y=1.02)
ticks=[0,8,16,24,32,40, 48]
plt.xticks(ticks, fontsize=7)
plt.yticks(ticks, fontsize=7)
plt.imshow(img, cmap='gray', interpolation='none')
# adding a red marker for highlighting the open
plt.plot(y, x, '.r', linewidth=0)
In [9]:
draw_group(images[100:116], centers[100:116])
## Class distribution (or class imbalance)¶
It is also highly recommended that you calculate the class distribution of your data. Try to count how many samples belong to each class. Like how many samples have an open defect at the point (32,15) (or in the region 17, if you go for the regions scheme). This is important to do in order to tune your expectations for model accuracy. The more the imbalance, the more chances that your model will be less accurate and biased. Therefore, you may need to carefully pick your training dataset from the above 400,000 samples. This task is of course part of this project ...
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2019-10-21 05:09:48
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https://physics.stackexchange.com/questions/454102/will-the-motion-in-both-systems-be-same/455683
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# Will the motion in both systems be same?
In my book, the discussion in SHM is mainly about a mass hanging vertically from a spring. But there are some exercises which contain questions which involves a mass attached to a spring and the whole system is horizontal.
So one system is vertical and another is horizontal. So won't $$g$$ affect them differently? Will their motion be same (like they will be in SHM but their $$\omega$$ will change)? Will equations like $$v=\omega \sqrt{A^2-x^2}$$ be valid for both of them?
The interesting thing is that the period of the motion will be the same in both cases but the horizontal spring must also obey Hooke's law in compression.
For the vertical spring let the unit vector in the down direction be $$\hat d$$.
Using Newton's second law for the static equilibrium position $$mg\, \hat d-kx_0 \,\hat d = 0\, \hat d \Rightarrow mg = kx_0$$.
Now consider the spring stretch an extra amount $$x$$ from its equilibrium position.
$$mg\, \hat d -k(x_0+x) \,\hat d = ma \,\hat d \Rightarrow -kx = ma$$ which is exactly the same equation of motion you would get for a horizontal spring with $$x$$ as the actual extension of the spring as for a horizontal spring the equilibrium position would be with the spring unextended.
Update in response to a question from the OP.
If the spring is horizontal and extended from its unstretched state in the $$\hat x$$ direction by an amount $$x$$ then applying Newton’s second law gives $$-kx \,\hat x = ma\,\hat x\Rightarrow -kx = ma$$.
If the spring-mass system is on a slope you will get the same equation of motion but a changed static equilibrium extension $$x_0$$.
Is is as though the gravitational field strength $$g$$ had been reduced.
• Can you please derive the equation of horizontal spring please? Why would it be the same? Will it be same regardless of the angle it makes with the base? – Theoretical Jan 14 at 13:32
Just read your post on Physics Meta. I think you should upvote the answers you like and search the site thoroughly before posting.
As for the question, you can prove this on your own. In the vertical spring case, first solve for the equilibrium case, then solve for the force after it is stretched by a small amount $$x$$. Equate the resultant force to $$ma$$ and you should get a an SHM equation which will be the same as what you would have gotten if you were solving it in the horizontal case.
The force of gravity does not affect the time period of the block because it is not a restoring force. It does not change it's magnitude with the position of the block.
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2019-10-18 19:13:03
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http://xiaoxu.lxxm.com/page/3/
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## Correction: the convenient radius for 95% confidence interval of t-test
-- What do you call a tea party with more than 30 people?
-- A Z party!!!
Joke #123 on http://www.ilstu.edu/~gcramsey/Gallery.html
2*SE is a popular convenient radius to eye 95% CI for t-test. Statisticians take t with df>=30 as z. However, I was incorrect to teach that 1.96*SE could be the precise radius when df>=30.
Guess what is the critical df to use decimally precise 1.96*SE. Larger than expected --
## R: str(…) 与 getS3method(…,…)
me: 请教两个R的技术：1.R中有没有对象浏览器之类的工具？一举看完一个对象的子子孙孙 2.怎么看深入的源代码> prcomp function (x, ...) UseMethod("prcomp") <environment: namespace:stats>
Yihui: 1. str()是很常用的一个函数，它可以充分查看对象的子子孙孙 2. 很多函数要么是S3 method，要么是调用C code，所以一般不能直接看源代码
S3 method可以用getS3method()去查看，比如prcomp就是S3方法，那么可以看它的default方法是什么：
> getS3method('prcomp','default')
function (x, retx = TRUE, center = TRUE, scale. = FALSE, tol = NULL, ...) { x <- as.matrix(x) x <- scale(x, center = center, scale = scale.) cen <- attr(x, "scaled:center") sc <- attr(x, "scaled:scale") if (any(sc == 0)) stop("cannot rescale a constant/zero column to unit variance") s <- svd(x, nu = 0) s$d <- s$d/sqrt(max(1, nrow(x) - 1)) if (!is.null(tol)) { rank <- sum(s$d > (s$d[1L] * tol)) if (rank < ncol(x)) { s$v <- s$v[, 1L:rank, drop = FALSE] s$d <- s$d[1L:rank] } } dimnames(s$v) <- list(colnames(x), paste("PC", seq_len(ncol(s$v)), sep = "")) r <- list(sdev = s$d, rotation = s$v, center = if (is.null(cen)) FALSE else cen, scale = if (is.null(sc)) FALSE else sc) if (retx) r$x <- x %*% s$v class(r) <- "prcomp" r } <environment: namespace:stats>
me: 多谢，节约俺好多搜索时间
Yihui: 嗯，我也是花了很长时间才明白S3 method的意思，呵呵
me: 如果要看biplot.prcomp呢？
Yihui: help会告诉你biplot也是generic function，可以应用在prcomp这种class上，所以：
> getS3method('biplot','prcomp') function (x, choices = 1:2, scale = 1, pc.biplot = FALSE, ...) { if (length(choices) != 2) stop("length of choices must be 2") if (!length(scores <- x$x)) stop(gettextf("object '%s' has no scores", deparse(substitute(x))), domain = NA) if (is.complex(scores)) stop("biplots are not defined for complex PCA") lam <- x$sdev[choices] n <- NROW(scores) lam <- lam * sqrt(n) if (scale < 0 || scale > 1) warning("'scale' is outside [0, 1]") if (scale != 0) lam <- lam^scale else lam <- 1 if (pc.biplot) lam <- lam/sqrt(n) biplot.default(t(t(scores[, choices])/lam), t(t(x\$rotation[, choices]) * lam), ...) invisible() } <environment: namespace:stats>
## Unexpectedly, the theoretically best reject-region of T-test is bounded.
$f_{t}\left(x;\mu,df\right)\equiv C\left(df\right)\left(1+\frac{\left(x-\mu\right)^{2}}{df}\right)^{-\frac{df+1}{2}}$
$\lambda\left(x;\mu_{0},\mu_{1},df\right)\equiv\frac{f_{t}\left(x;\mu_{1},df\right)}{f_{t}\left(x;\mu_{0},df\right)}=\left(\frac{v+\left(x-\mu_{1}\right)^{2}}{v+\left(x-\mu_{0}\right)^{2}}\right)^{-\frac{df+1}{2}}{\longrightarrow\atop x\rightarrow\infty}1$
For NHST $H_{0}:T\sim t_{df}$ vs $H_{1}:T-1\sim t_{df}$, theoretically, $p\left(t\right)=\int_{\left\{ x:\lambda\left(x\right)\ge\lambda\left(t\right)\right\} }f_{t}\left(x,\mu_{0},df\right)dx$ is s.t. $\lim_{t\rightarrow\infty}p\left(t\right)=\frac{1}{2}$ , rather than zero. Nevertheless, pratically a large t, rejecting both $H_{0}$ and $H_{1}$, should not be counted as any evidence to retain or reject $H_{0}$.
To verify the shape of $\lambda\left(x\right)$ --
## Confidence Region and Not-reject Region
Either Confidence Interval (CI) or Null Hypothesis Significance Test (NHST) has the same business, to advise whether some sample $X\equiv\left(X_{1},X_{2},\dots,X_{n}\right)$ is or is not disliked by some hypothesized parameter $\vartheta$.
NHST.com manages a database. For each Miss $\vartheta$, NHST spies out all she dislikes. Mr X logs in NHST.com and inputs a girl name and his credit card number, to bet his luck whispering-- Does she dislike me?
CI.com manages a database too. For each Mr X, CI only needs his credit card with his name X on it, then serves him a full list of available girls.
NHST.com has been historically monopolizing the market. Nevertheless, somebody prefer visiting CI.com and find that the two may share database in most cases.
Not-reject Region of $\vartheta$ is defined as $A\left(\vartheta\right)=\left\{ x:\vartheta\; doesn't\;dislike\;x\right\}$.
Confidence Region of x is defined as $S\left(x\right)\equiv\left\{ \vartheta:\vartheta\; doesn't\;dislike\;x\right\}$.
$\theta\in S\left(X\right)\Leftrightarrow \theta\,does\,not\,dislike\,X$ $\Leftrightarrow\,X\in\,A\left(\theta\right)$
So, $Pr_{\vartheta}\left(\vartheta\in S\left(X\right)\right)\ge1-\alpha,\forall\vartheta\Longleftrightarrow Pr_{\vartheta}\left(X\notin A\left(\vartheta\right)\right)\le\alpha,\forall\vartheta$
## Automatize LISREL jobs
LISREL routine can run in DOS or in command line mode of windows (windows-key + R -> CMD) . The command line is just like --
D:\My Documents>"C:\Program Files\lisrel87\lisrel87.exe" "C:\Program Files\lisrel87\LS8EX\EX61.LS8" D:\myOutput.out
1. You only need edit and input the bold part.
2. Quotation marks are used wherever the paths or filenames include blanks.
3. The 2nd argument is the output file. You can still specify more output options in your .ls8 file.
4. A .bat file can automatize batches of such lisrel jobs.
## Developing normal pdf from symmetry & independence
When I was in the 3rd grade of my middle school, I enjoyed my town bookstore as a standing library. There a series of six math-story books by Zhang Yuan-Nan impressed me a lot. I cited a case from one in my PPT when I taught the normal distribution -- the normal pdf can be derived from simple symmetry & independence conditions.
Today I can even google out an illegal pdf of its new edition to verify the case (2005, pp. 89). Actually I have bought the new edition series (now 3 books) and lent them to students. Those conditions are as instinctive as--
1. For white noise errors on 2-D, the independence means pdf at $(x,y)$ is the product of 1-D pdf, that is, $\phi\left(x\right)\phi\left(y\right)$.
2. The symmetry means pdf at $(x,y)$ is just a function of $x^{2}+y^{2}$, nothing to do with direction. That is, $\phi\left(x\right)\phi\left(y\right)=f\left(x^{2}+y^{2}\right)$.
So, $f\left(x^{2}\right)f\left(y^{2}\right)=f\left(x^{2}+0\right)f\left(0+y^{2}\right)=\phi^{2}\left(0\right)f\left(x^{2}+y^{2}\right)$.
For middle school students, the book stated a gap here to arrive at the final result $f\left(x^{2}\right)=ke^{bx^{2}}$, which is $\phi\left(x\right)=\frac{1}{\phi\left(0\right)}f\left(x^{2}+0\right)=\frac{k}{\phi\left(0\right)}e^{bx^{2}}$.
I think non-math graduate students with interests can close the gap by themselves with following small hints.
Denote $\alpha=x^{2},\beta=y^{2}$.
We have
$\log f\left(\alpha\right)+\log f\left(\beta\right)=\log\phi^{2}\left(0\right)+\log f\left(\alpha+\beta\right)$,
or
$\;\;\left[\log f\left(\alpha\right)-\log\phi^{2}\left(0\right)\right]+\left[\log f\left(\beta\right)-\log\phi^{2}\left(0\right)\right]$ $=\left(\log f\left(\alpha+\beta\right)-\log\phi^{2}\left(0\right)\right)$.
Denote $g\left(\alpha\right)=\log f\left(\alpha\right)-\log\phi^{2}\left(0\right)$.
That is, $g\left(\alpha\right)+g\left(\beta\right)=g\left(\alpha+\beta\right)$.
Now to prove $g\left(\frac{m}{n}\right)=\frac{m}{n}g\left(1\right),\forall m,n\in\mathbb{N}$. With continuousness, it gets $g\left(\alpha\right)=\alpha g\left(1\right),\forall\alpha\in\mathbb{R}$.
## [二度更新并推荐]愉快地发现SciTE和LyX在WinXP下都支持中文
SciTE的设置是Options->Open Global Options File，编辑SciTEGlobal.properties，找到如下段落
# Unicode
#code.page=65001
code.page=0
#character.set=204
# Required for Unicode to work on GTK+:
#LC_CTYPE=en_US.UTF-8
#output.code.page=65001
# Unicode
code.page=65001
#code.page=0
character.set=204
# Required for Unicode to work on GTK+:
LC_CTYPE=en_US.UTF-8
output.code.page=65001
[update] 除了utf-8, SciTE 还支持国内更常用的GBK码，设置如下：
code.page=936
output.code.page=936
character.set=134
LyX(版本>=1.5.1)在winXP已经可以在.lyx文件正文和公式框中录入中文。麻烦的是输出中文的pdf。[UPDATED update]LyX的最新版本(1.6.2)捆绑MikTeX的安装包已经对中文(unicode)支持得很好了。感谢楼下joomlagate先生email给我的情报：http://cohomo.blogbus.com/logs/31361739.html 的后半篇介绍了通过XeTeX输出pdf的简单设置。我今天试了一下，效果非常理想。
[update]公式框中的中文只需要再ctrl-M一次即可。例如，\frac{\mbox{分子}}{\mbox{分母}}可以输出，而\frac{分子}{分母}不行。
LyX主页 http://lyx.org/
LyX中设置XeTeX中文支持的介绍： http://cohomo.blogbus.com/logs/31361739.html
## My first wp plugin work: LaTeX_Math_cgi 1.0
Activate it in Plugins menu. View its options in Options > LaTeX tab --
It is actually used as a mimeTeX plugin rather than just a $L^{A}T_{E}X$ plugin. My contribution is technically trivial. But, you must need it if you change from a light theme to a cool black one while find the default mimeTeX images are black in front and transparent in background. This plugin provides mimeTeX's \reverse and \opaque options to tune your math forms to your wp theme without editing them one by one.
The 2nd function is the option for your cgi URL. The default is http://www.forkosh.dreamhost.com/mimetex.cgi You can DIY one following Forkosh's instructions and then set it like /cgi-bin/mimetex.cgi It can help your visitors not to cost Forkosh's unselfish bandwidth.
## Understanding QQ plots
## Try distributions like rchisq, rt, runif, rf to view its heavy, or light, left, or right tail.
n <- 30; ry <- rnorm(n); qqnorm(ry);qqline(ry); max(ry) min(ry) ##view and guess what are x(s) and y(s) I <- rep(1,n); qr <- ((ry%*%t(I) > I %*% t(ry))+.5*(ry %*% t(I) == I%*%t(ry)))%*%I *(1/n);##qr are the sample quantiles points(qr,ry,col="blue"); ##to view the fact, try the following points(qr,qr*0,col="green",pch="|"); rx <- qnorm(qr); points(rx,ry,col="red",pch="O"); ##Red O(s) circle black o(s) exactly.
## 03DEC2007 R-workshop sponsored by dept of psy, ZSU(=SYSU, Guang-Zhou)
Here is the updated PPT for the talk in the afternoon--which includes the zipped example codes and set-up steps for the workshop in the evening within the 3rd page. The listed anonymous on-line test (result statistics) on p-value interpretation was cited indirectly From Gigerenzer, Krauss, & Vitouch (2004).
There is an advert on http://www.psy.sysu.edu.cn/detail_news.asp?id=258 and a formal CV of the speaker is available on http://lixiaoxu.googlepageS.com
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2017-08-22 03:42:26
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https://www.ques10.com/p/36029/explain-the-terms-dbt-wbt-and-rh/
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0
7.8kviews
Explain the terms DBT, WBT and RH
Subject:- Refrigeration and Air Conditioning
Topic:- Design of air conditioning systems
Difficulty:- Low
0
62views
Dry bulb temperature (DBT or $t_d$):
The temperature of moist air as measured by ordinary thermometer when placed in air is called dry bulb temperature.
Wet bulb temperature (WBT or $t_w$):
It is the temperature recorded by a thermometer whose bulb is covered with a wick or cloth saturated with water and is exposed to a current of moving air. When air is passed over the wet wick, moisture contained in the wick evaporates and cooling effect is produced at the bulb. Hence WBT is always less than DBT for unsaturated air and is equal to DBT for saturated air. WBT value is affected by moisture content of air.
The difference between DBT and WBT is known as Wet Bulb Depression (WBD) and is measure of degree of saturation of air. WBD=DBT-WBT
It is defined as the ratio of actual mass of water vapour $m_v$ in a certain sample of moist air at temperature $‘t_d’$ to the mass of water vapour in the same volume of saturated air $m_vs$ at same temperature $‘t_d’$ and same barometric pressure.
$R.H.=φ=\frac{m_v}{m_vs} =\frac{p_v}{p_vs}$
Relative Humidity can also be defined as the ratio of actual partial pressure of water vapour in unsaturated air to maximum partial pressure of water vapour at the same temperature when air becomes saturated. It is expressed in %. For saturated air R.H.= 100%.
Specific Humidity: Specific or absolute humidity or humidity ratio or moisture content is defined as the ratio of mass of water vapour to the mass of dry air in a given sample of air.
$ω=\frac{m_v}{m_a}$
Unit of specific humidity
$ω=\frac{kg of water vapour}{kg of dry air}$
Degree of Saturation: It is defined as the ratio of specific humidity for a given sample of air to the specific humidity of saturated air at same dry bulb temperature and same barometric pressure.
$μ=\frac{ω}{ω_s}$
For dry air, $μ=0(as ω=0)$
For saturated air, $μ=1(asω=ω_s)$
Dew Point Depression- It is the difference between the dry bulb temperature and dew point temperature of air.
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2021-06-12 11:15:50
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http://www.physicsforums.com/showthread.php?p=4271512
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## Log expansion for infinite solenoid
Hello, I found an approximation for this log function:
$$log \Bigg(\frac{\Lambda}{\rho} + \sqrt{1 + \frac{\Lambda^2}{\rho^2}} \Bigg),$$
where $\Lambda \rightarrow \infty$. The above is approximated to the following,
$$-log \bigg(\frac{\rho}{\rho_o} \bigg) + log \bigg(\frac{2 \Lambda}{\rho_o} \bigg).$$
How is this done? I tried expanding the $\sqrt{1 + x^2}$ term, but I still don't get how they arrive to the above approximation.
Any help would be greatly appreciated!
Cheers!
I have no idea why this was sent to linear algebra section . . . And I do not know how to move it to classical physics. . .
PhysOrg.com science news on PhysOrg.com >> New language discovery reveals linguistic insights>> US official: Solar plane to help ground energy use (Update)>> Four microphones, computer algorithm enough to produce 3-D model of simple, convex room
Blog Entries: 1 Recognitions: Gold Member Homework Help Science Advisor What is ##\rho_0##? It appears in the second expression but not the first.
I do not know what $\rho_o$ is, I assume some constant. I found this approximation here: http://www.physicsforums.com/showthread.php?t=119419
## Log expansion for infinite solenoid
Wow, never mind. Clearly I am being silly here, for $\Lambda \rightarrow \infty$.
$$log\bigg( \frac{\Lambda}{\rho} + \sqrt{1 + \frac{\Lambda^2}{\rho^2}} \bigg) \rightarrow log \bigg( \frac{ 2 \Lambda}{\rho} \bigg) \rightarrow log(2 \Lambda) - log(\rho).$$
As for the $\rho_o$ I have no idea why that enters the equation.
Tags expansion, infinite solenoid, magnetostatics
Similar discussions for: Log expansion for infinite solenoid Thread Forum Replies Introductory Physics Homework 1 Introductory Physics Homework 3 General Physics 10 Advanced Physics Homework 3 Classical Physics 17
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2013-06-19 01:35:07
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https://arxiver.moonhats.com/2017/12/07/relation-between-parameters-of-dust-and-parameters-of-molecular-and-atomic-gas-in-extragalactic-star-forming-regions-ga/
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# Relation between parameters of dust and parameters of molecular and atomic gas in extragalactic star-forming regions [GA]
The relationships between atomic and molecular hydrogen and dust of various sizes in extragalactic star-forming regions are considered, based on observational data from the Spitzer and Herschel infrared space telescopes, the Very Large Array (atomic hydrogen emission) and IRAM (CO emission). The source sample consists of approximately 300 star-forming regions in 11 nearby galaxies. Aperture photometry has been applied to measure the fluxes in eight infrared bands (3.6, 4.5, 5.8, 8, 24, 70, 100, and 160$\mu$m), the atomic hydrogen (21cm) line and CO (2–1) lines.
The parameters of the dust in the starforming regions were determined via synthetic-spectra fitting, such as the total dust mass, the fraction of polycyclic aromatic hydrocarbons (PAHs), etc. Comparison of the observed fluxes with the measured parameters shows that the relationships between atomic hydrogen, molecular hydrogen, and dust are different in low- and high-metallicity regions. Low-metallicity regions contain more atomic gas, but less molecular gas and dust, including PAHs. The mass of dust constitutes about $1\%$ of the mass of molecular gas in all regions considered. Fluxes produced by atomic and molecular gas do not correlate with the parameters of the stellar radiation, whereas the dust fluxes grow with increasing mean intensity of stellar radiation and the fraction of enhanced stellar radiation. The ratio of the fluxes at 8 and 24$\mu$m, which characterizes the PAH content, decreases with increasing intensity of the stellar radiation, possibly indicating evolutionary variations of the PAH content. The results confirm that the contribution of the 24$\mu$m emission to the total IR luminosity of extragalactic star-forming regions does not depend on the metallicity.
K. Smirnova, M. Murga, D. Wiebe, et. al.
Thu, 7 Dec 17
57/72
Comments: Published in Astronomy Reports, 2017, vol. 61, issue 8
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2017-12-11 11:38:02
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|
https://www.techwhiff.com/issue/what-is-one-reason-scholars-believe-that-public-opinion--493822
|
# What is one reason scholars believe that public opinion matters in american politics?
###### Question:
What is one reason scholars believe that public opinion matters in american politics?
### Is this right? Im trying to finish all of my work!!
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### Scores on the SAT college entrance test in a recent year were roughly Normal with mean 1012.5 and standard deviation 184.9. You choose an SRS of 90 students and average their SAT scores. If you do this many times, the mean of the average scores you get will be close to
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### Hey help a tom outtttttttttttttttttttttt
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### 10. Which property is used to show the form in maximize state a) Selection State b) Window State c) Mode d) Opening State
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### In what ways was the Ottoman Empire influenced both by previous Islam empires and the Byzantine Empire?
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### Name the six animals found in the movie trailer a cry in the wild
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### PLEASE STOP GIVING ME FAKE ANSWERS!!!!! >:( LOOK AT THIS QUESTION THAT I POSTED EARLIER TODAY. THIS PERSON ANSWERED "B" BUT I DIDN'T GIVE YOU ANY ANSWER CHOICES. YOU DON'T MAKE SENSE. im about to delete this app cuz no one seems to care. IF YOU DONT KNOW THE ANSWER, THE. STOP MAKING A FOOL OUT OF YOURSELVES AND DONT ANSWER. jeez.
PLEASE STOP GIVING ME FAKE ANSWERS!!!!! >:( LOOK AT THIS QUESTION THAT I POSTED EARLIER TODAY. THIS PERSON ANSWERED "B" BUT I DIDN'T GIVE YOU ANY ANSWER CHOICES. YOU DON'T MAKE SENSE. im about to delete this app cuz no one seems to care. IF YOU DONT KNOW THE ANSWER, THE. STOP MAKING A FOOL OUT OF...
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### The story ,( The girl who couldn't see herself. )Does the mood of the story improves gradually?
The story ,( The girl who couldn't see herself. )Does the mood of the story improves gradually?...
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2023-03-23 18:27:11
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https://cs.stackexchange.com/questions/79264/basic-mapping-reductions-without-using-turing-machines
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# Basic mapping reductions without using Turing machines
I have problems with the basics of mapping reductions.
I can understand how to do reductions using a Turing machine, but without it, I get a little bit confused.
For example:
1. How do I do a mapping reduction from $a$ to $b$, where $a$ is $\{0,1\}$ and $b$ is $\{0\}$ over the alphabet of $\{0,1\}$?
2. Does every language have a reduction to itself?
The identity function is a reduction from a language to itself. One reduction from $\{0,1\}$ to $\{0,1\}$ is given by $\lfloor x/2 \rfloor$, which maps $0,1$ to $0$, and all other integers to non-zero values.
• @aradona A suitable reduction would be a function $f$ which is defined to map 0 and 1 to 0 and any other input to 1. – Rick Decker Jul 24 '17 at 14:12
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2020-05-27 14:54:51
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https://physics.stackexchange.com/questions/460211/merging-black-holes-how-fast-do-they-hit
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# Merging black holes — how fast do they hit?
Orbital mechanics says something falling from infinity hits at the escape velocity of the object it strikes.
The escape velocity at the event horizon is lightspeed, it must be higher for the singularity inside.
Thus we have a superluminal impact velocity. Huh? Does special relativity manage to avoid this situation?
• You might be referring to the "pseudo-derivation" of the Schwarzschild radius of a black hole. You can treat it like a classical mechanics problem of a light-speed escape velocity, but the actual physics is more complicated than this. It is just a coincidence that using the "pseudo-derivation" still gives Schwarzschild radius – Aaron Stevens Feb 11 at 18:46
• From an outside observer, it appears they never hit due to time dilation. But then how do we get GRB and other observations? – ja72 Feb 11 at 19:55
• Newtonian orbital mechanics says that. Special relativity avoids the situation by not handling gravitation at all. General relativity could handle it (and does, numerically), but it is very very complicated and there are no equations for merging black holes. – m4r35n357 Feb 11 at 20:38
The OP could be asking either how fast the singularities hit or how fast the event horizons hit.
Singularities
The singularities are spacelike, so it's not well defined to ask about their velocity. The velocity of A relative to B is only something we expect to be well defined if A and B are timelike or lightlike world lines (or surfaces with similar properties). A singularity is basically the end of the world for observers inside the horizon, so asking how fast one singularity is moving relative to another is like asking how fast the world is moving relative to someone else's end of the world. It doesn't make sense. The end of the world is a time, not an object.
Event horizons
GR doesn't have global frames of reference or a way to measure the velocity of an object relative to a distant observer. See How do frames of reference work in general relativity, and are they described by coordinate systems? . So there is not any meaningful sense in which we can talk about the velocity of an event horizon as seen by some distant observer, only by an observer near the horizon. So this means that if we were hoping to get an answer about the speed at which the event horizons collide as they merge, it would have to be the speed as measured by an observer right near the point of merging.
An event horizon is a lightlike surface, so in general all observers near a horizon say that a horizon is moving at the speed of light. So the observer probably sees both horizons as moving at $$c$$ as they collide. This doesn't mean that one horizon is moving at $$2c$$ relative to the other horizon, because we don't have frames of reference moving at $$c$$, i.e., there can't be an observer at rest relative to one of the horizons.
• While you can't have an observer sitting on the event horizon you could have one in orbit about one of the holes orbiting perpendicular to the axis of the collision. I do agree that c + c doesn't give 2c but even at the event horizon we should see c--yet Einstein says we can't. – Loren Pechtel Feb 12 at 5:22
• @LorenPechtel: While you can't have an observer sitting on the event horizon you could have one in orbit about one of the holes To be in orbit, you have to be at a radius greater than or equal to 3 times the Schwarzschild radius $r$ (the innermost stable circular orbit, or ISCO). So you're at a significant distance from the horizon (significant compared to $r$, which sets the scale of curvature), and therefor you don't have any meaningful way of talking about the horizon's velocity relative to you. This is the idea of the linked Q&A. – Ben Crowell Feb 12 at 19:59
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2019-02-21 19:44:36
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https://solvedlib.com/n/reaction-mechanisms-the-reaction-between-mathrm-no-and-mathrm,11883738
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# Reaction Mechanisms The reaction between $\mathrm{NO}$ and $\mathrm{Cl}_{2}$ is first order in each reactant. Does this mean that the reaction
###### Question:
Reaction Mechanisms The reaction between $\mathrm{NO}$ and $\mathrm{Cl}_{2}$ is first order in each reactant. Does this mean that the reaction could occur in just one step? Explain your answer.
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##### Question 1:(2 marks) Find the domain and the range of each of the following functionsFind the domain and range ofthe function (1 point) Find the domain and range ofthe function y -2 (1 point)Question 2:12 marks) Find the equation of the line that passes through the points (1,4) and (-3,2)Question 3(10 marks) For cach of the following functions find the derivative from first principles and clearly demonstrate all steps. f() = 2x + 3f6x) =4f(x) = Vxf(x) = 3x2 _ 8fx) = 9x"
Question 1: (2 marks) Find the domain and the range of each of the following functions Find the domain and range ofthe function (1 point) Find the domain and range ofthe function y -2 (1 point) Question 2: 12 marks) Find the equation of the line that passes through the points (1,4) and (-3,2) Quest...
##### Question 10 0f 17Reaction BJacate CehyoropenaseH;C-OHNADHDraw the product of Reaction €Rcaction € .Mae canucreneNADH
Question 10 0f 17 Reaction B Jacate Cehyoropenase H;C- OH NADH Draw the product of Reaction € Rcaction € . Mae canucrene NADH...
##### 9. (15 points) Data are beiag collected from metal stamping process where presses are simultaneously making metal clips that will be used in the assembly of automobile instrument panels. The quality characteristic is an inclination angL of the clip_ The specs based on the angle are 15 = 2.5 degrees. Statistical charting has been initiated on this process and the sample size is 4. The averages and ranges of 40 subgroups are given in the table below. Construct the mean and range charts and interpr
9. (15 points) Data are beiag collected from metal stamping process where presses are simultaneously making metal clips that will be used in the assembly of automobile instrument panels. The quality characteristic is an inclination angL of the clip_ The specs based on the angle are 15 = 2.5 degrees....
##### Application Problem 6-11A a-c DejaVu Company has been in business for several years and has the...
Application Problem 6-11A a-c DejaVu Company has been in business for several years and has the following information for its operations in the current year: Total credit sales Bad debts written off in the year Accounts receivable balance on December 31 (after writing off the bad debts above) \$3,186...
##### Conslccr Ihcac compoundsCrOH) AzPo4 PbS CaSo4Complcte thc following statcments by cntering thc letter(s) corresponding the correct compound(s) (If morc than onc compound fits thc dcscriptionincludc all the relevant compounds by writing Your answcr chuicict; H thout pucmuanion cgABC )string ofWuthou dongcalculations it is possible dctcrminc that nickel(l) sullide Morc soluble thanand nickel(ll) sulfide less soluble thznIt is po- possiblc = detcrine whcther nickel(TI) sulfide mcneless soluble tha
Conslccr Ihcac compounds CrOH) AzPo4 PbS CaSo4 Complcte thc following statcments by cntering thc letter(s) corresponding the correct compound(s) (If morc than onc compound fits thc dcscriptionincludc all the relevant compounds by writing Your answcr chuicict; H thout pucmuanion cgABC ) string of Wu...
##### Purified Spectraeach spectruim, draw the compound and designate which peaks = coordinate t0 which ltures on your compound Include the spectra with your report How might the aspirin cctra appear if it was not completely purified (ie. if there was left over salicylic acid)?'H-NMR spectrum for salicylic acidMacBook Pro838% 5&8
Purified Spectra each spectruim, draw the compound and designate which peaks = coordinate t0 which ltures on your compound Include the spectra with your report How might the aspirin cctra appear if it was not completely purified (ie. if there was left over salicylic acid)? 'H-NMR spectrum f...
##### How do you solve -6( 8+ 4a ) > 32- 8a?
How do you solve -6( 8+ 4a ) > 32- 8a?...
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2023-03-29 16:07:04
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http://stackoverflow.com/questions/11410562/how-to-control-wrong-pagebreaks-of-longtable-in-latex-from-hmisc-package
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# How to control wrong pagebreaks of longtable in latex() from Hmisc package?
I'm using Sweave and latex() from the Hmisc package to insert a longtable in my PDF.
When I do it the first time, the table is spread nicely, filling up the pages with the table.
If I do it again, some pages are only half full (like page 4 of PDF), which looks weird and somehow wrong because it seems to be unnecessary space.
Is there a way to control this? Or what could I do to improve the look? Especially, if I'm adding some text and graphs, it won't look good with the empty space on page 4.
\documentclass{article}
\usepackage{Sweave}
\usepackage{longtable}
\usepackage{booktabs}
\begin{document}
\SweaveOpts{concordance=TRUE}
I want to insert this longtable
<<tab.R,echo=FALSE,results=tex>>=
library(Hmisc)
#library(xtable)
x <- matrix(rnorm(1000), ncol = 10)
x.big <- data.frame(x)
latex(x.big,"",file="",longtable=TRUE, dec=2,caption='First longtable spanning several pages')
@
then write some text. Maybe add a graph...
And then another table
<<tab.R,echo=FALSE,results=tex>>=
latex(x.big,"",file="",longtable=TRUE, dec=2,caption='Second longtable spanning wrongly')
@
\end{document}
-
you know that there exists a stackexchange portal dedicated to latex questions - it is to be found at tex.stackexchange.com. I have flagged your question for migration. – epsilonhalbe Jul 10 '12 at 11:45
Actually, I didn't know, I'm quite new. Thanks for pointing that out! By flagging it for migration, will it automatically appear on tex.stackexchange.com or do I have to ask again there? – physh Jul 10 '12 at 15:18
it can take some time but i think it will be migrated, as soon as some moderator looks this way. – epsilonhalbe Jul 10 '12 at 16:36
Don't pass this question over to the latex group, this is a problem of Hmisc/latex that adds a \clearpage into tex every 40 lines by default. Check parameter lines.page=40 of latex. I do not understand why this default has been set, but something like
latex(x.big,"",file="",longtable=TRUE, dec=2,
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2013-12-06 14:41:49
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https://www.genetics.org/highwire/markup/391217/expansion?width=1000&height=500&iframe=true&postprocessors=highwire_tables%2Chighwire_reclass%2Chighwire_figures%2Chighwire_math%2Chighwire_inline_linked_media%2Chighwire_embed
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TABLE 4
Cross-validation analysis of QTL detection and comparison to whole data (WD) set results
Conventional F3LHRF-F3
DGYGMSDDGYGMSD
QTL no.
WD99121155
DS8.966.008.315.683.755.93
Range(5–15)(3–10)(1–14)(1–11)(1–11)(2–12)
WD48.40/59.7545.50/55.4945.70/54.4041.50/56.8520.60/24.8226.00/30.23
DS54.03/66.7042.29/51.5741.02/48.8330.58/41.8921.4/25.7833.86/39.37
VS39.48/48.7431.05/37.8720.05/23.8712.24/16.766.05/7.2917.22/20.02
Bias1.371.362.042.503.541.97
Av. R2 per QTL4.4/5.45.2/6.32.4/2.92.2/2.91.6/1.92.9/3.4
• The number of detected QTL (QTL no.) and the proportion of phenotypic and genotypic variance explained by the detected QTL () are reported for each analysis. In the cross-validation process, the number of detected QTL corresponds to the average number of detected QTL across 200 5-split cross-validation detection sets (DS). We also indicated the range of variation of the number of QTL detected (range).Two different types of percentages of variance are given: the average values across detection sets (DS) and validation sets (VS). The ratio between the DS value and the VS value gives an estimate of the selection bias for the percentage of variance explained. The VS values were divided by the average number of QTL detected in DS to estimate the average contribution of each individual QTL to the variation (Av. R2 per QTL).
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2021-06-21 23:38:20
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https://www.shaalaa.com/question-bank-solutions/find-value-k-which-each-following-system-equations-has-infinitely-many-solutions-x-k-1-y-4-k-1-x-9y-5k-2-pair-linear-equations-two-variables_22411
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# Find the Value Of K For Which Each of the Following System of Equations Has Infinitely Many Solutions : X + (K + 1)Y =4 (K + 1)X + 9y - (5k + 2) - Mathematics
Find the value of k for which each of the following system of equations has infinitely many solutions :
x + (k + 1)y =4
(k + 1)x + 9y - (5k + 2)
#### Solution
The given system of the equation may be written as
x + (k + 1)y - 4 = 0
(k + 1)x + 9y - (5k + 2) = 0
The system of equation is of the form
a_1x + b_1y + c_1 = 0
a_2x + b_2y + c_2 = 0
Where a_1 = 1,b_1 = k + 1, c_1 = -4
And a_2 = k +1, b_2 = 9, c_2 = -(5k + 2)
For a unique solution, we must have
a_1/a_2 = b_1/b_2 = c_1/c_2
=> 1/(k +1) = (k +1)/9 = (-4)/(-(5k + 2))
=> 1/(k +1) = (k +1)/9 and (k+1)/9 = 4/(5k +2)
=> 9 = (k +1)^2 and (k +1)(5k + 2) = 36
=> 9 = k^2 + 1 + 2k and 5k^2 + 2k + 5k + 2 = 36
=> k^2 + 2k + 1 - 9 = 0 and 5k^2 + 7k + 2 - 36 = 0
=> k^2 + 2k - 8 = 0 and 5k^2 + 7k - 34 = 0
=> k^2 + 4k -2k - 8 = 0 and 5k^2 + 17k - 10k - 34 = 0
=> k(k + 4) - 2(k + 4) = 0 and (5k + 17) - 2(5k + 17) = 0
=> (k + 4)(k -2) = 0 and (5k + 1) (k -2) = 0
=> (k = -4 or k = 2) and (k = (-17)/5 or k = 2)
k = 2 satisfies both the conditions
Hence, the given system of equations will have infinitely many solutions, if k = 2
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 10 Maths
Chapter 3 Pair of Linear Equations in Two Variables
Exercise 3.5 | Q 15 | Page 73
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2021-03-09 04:44:41
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http://tex.stackexchange.com/questions/59001/how-to-make-pgfgantt-scale-to-specific-widths-in-the-page-ex-textwidth
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# How to make pgfgantt scale to specific widths in the page? (ex. textwidth)
I'm trying to use pgfgantt to create a gantt chart for a project spanning many weeks, thus the following code would overflow through the right margin. I can't find a way to alter the size of the grid elements or time slots:
\begin{tikzpicture}[x=.5cm, y=1cm]
\begin{ganttchart}%
[vgrid, hgrid]{50} % 50 weeks
\gantttitle{Title}{12} \\
\ganttbar{Task 1}{1}{4} \\
\ganttbar{Task 2}{5}{6} \\
\ganttmilestone{M 1}{6} \\
\ganttbar{Task 3}{7}{11}
\ganttlink{4}{2}{5}{3}
\ganttlink[b-m]{6}{3}{6}{4}
\ganttlink[m-b]{6}{4}{7}{5}
\end{ganttchart}
\end{tikzpicture}
My former question above, however here's a new angle...
How can I make pgfgantt scale everything by limiting its size to specific parameters? ex. I want it to span the full textwidth of the page or the full width save for the margins.
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Could you turn your code snippet into a compilable document by adding the necessary preamble and \begin{document}? – Jake Jun 8 '12 at 7:05
## 3 Answers
With command \resizebox{new width}{new height}{what you want to resize} you can change dimensions of any latex box. This means that everything (text, lines, simbols, ...) will be adjusted to new dimensions. In first or second parameter you can use ! in order to keep original aspect ratio. With this command,
\documentclass{article}
\usepackage{lipsum}
\usepackage{pgfgantt}
\begin{document}
\lipsum[1]
\noindent\resizebox{\textwidth}{!}{
\begin{tikzpicture}[x=.5cm, y=1cm]
\begin{ganttchart}%
[vgrid, hgrid]{50} % 50 weeks
\gantttitle{Title}{12} \\
\ganttbar{Task 1}{1}{4} \\
\ganttbar{Task 2}{5}{6} \\
\ganttmilestone{M 1}{6} \\
\ganttbar{Task 3}{7}{11}
%\ganttlink{4}{2}{5}{3}
%\ganttlink[b-m]{6}{3}{6}{4}
%\ganttlink[m-b]{6}{4}{7}{5}
\end{ganttchart}
\end{tikzpicture}
}
\lipsum[2]
\end{document}
your Gantt chart would look something like:
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[fr] tu peux jouer sur l'échelle de la figure s'il y a des noeuds, il faudra aussi ajouter transform shape mais cela risque de ne plus être très joli
[en] you can play on the global scale of the figure if there are knots, it will also add transform shape but this may not be very pretty
\documentclass[A4]{article}
\usepackage{tikz}
\usepackage{lipsum}
\begin{document}
\noindent \begin{tikzpicture}[scale=\textwidth/8cm]
\draw (0,0) rectangle (8,3);
\end{tikzpicture}
\lipsum[1]
\noindent \begin{tikzpicture}[scale=\textwidth/8cm,transform shape]
\node [rectangle,minimum width=8cm,minimum height=3cm,draw]{aaa};
\end{tikzpicture}
\end{document}
-
One partial solution seems to use x unit and y unit to shrink the size of the time slots. This solves my first question.
I will wait for any potential answer for my second question. :)
-
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2015-02-28 06:56:30
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https://brilliant.org/discussions/thread/blink-fast/
|
×
...literally...
Note by Llewellyn Sterling
2 years, 6 months ago
Sort by:
Nice one !!
- 2 years, 6 months ago
Thanks,
- 2 years, 4 months ago
The secret of creating the animation, please?
- 2 years, 6 months ago
Well, there isn't a secret to be honest.
- 2 years, 4 months ago
Literally awesome!!!!!!!!!
- 2 years, 4 months ago
Thanks.
- 2 years, 4 months ago
Staring at this with or without glasses look totally cool.
- 2 years, 6 months ago
I know. However, it is little blurry without your glasses at times. :D
- 2 years, 4 months ago
No offence but actually there's no need to blink hard . Just almost close your eyes and look at the pic , to get the same effect :)
- 2 years, 6 months ago
For those wearing glasses, just remove them.
- 2 years, 6 months ago
Do you wear glasses ?
- 2 years, 6 months ago
I do and I'm the one that posted this. xD
- 2 years, 6 months ago
Yep! Degree 500 and 600
- 2 years, 6 months ago
Is that a GIF? Whatever it is, It is great!!
- 2 years, 6 months ago
Seems to be a GIF...
- 2 years, 6 months ago
Thanks and gif. :)
- 2 years, 4 months ago
Wow
- 2 years, 2 months ago
Pretty awesome, huh?
- 2 years, 2 months ago
Btw where is this gif found?
- 2 years, 4 months ago
In all fair honesty...I forgot how I got this gif. This was a few months ago.
- 2 years, 4 months ago
Nice
- 2 years, 6 months ago
Thank you.
- 2 years, 4 months ago
That's cool
- 2 years, 6 months ago
Thanks.
- 2 years, 4 months ago
Yahoo!
- 2 years, 6 months ago
- 2 years, 6 months ago
Haha!!!!!
- 2 years, 6 months ago
Bing! :D
- 2 years, 4 months ago
Awesome!!
- 2 years, 6 months ago
- 2 years, 4 months ago
Whoo!
- 2 years, 6 months ago
Fascinating to see, isn't it?
- 2 years, 4 months ago
It's really fascinating!
- 2 years, 4 months ago
I know; staring at this for a few minutes is quite interesting to be honest. Give it a try.
- 2 years, 4 months ago
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2017-10-17 15:18:58
|
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https://zbmath.org/?q=an:0554.60077
|
# zbMATH — the first resource for mathematics
Approximation by Brownian motion for Gibbs measures and flows under a function. (English) Zbl 0554.60077
Let $$\{S_{\tau}:$$ $$\tau\in {\mathbb{R}}\}$$ denote a flow built under a Hölder-continuous function l over the base ($$\Sigma$$,$$\mu)$$ where $$\Sigma$$ is a topological Markov chain and $$\mu$$ some ($$\psi$$-mixing) Gibbs measure. For a certain class of functions f with finite $$2+\delta$$- moments it is shown that there exists a Brownian motion B(t) with respect to $$\mu$$ and $$\sigma^ 2>0$$ such that $$\mu$$-a.e. $\sup_{0\leq u<l(x)}| \int^{t}_{0}fS_{\tau}d\tau -B(\sigma^ 2t)| \ll t^{-\lambda}\quad for\quad some\quad 0<\lambda <\delta /588.$ One can also approximate in the same way by a Brownian motion $$B^*(t)$$ with respect to the probability $$(\int l du)^{-1}l d\mu$$. From this, the central limit theorem, the weak invariance principle, the law of the iterated logarithm and related probabilistic results follow immediately. In particular, the result of M. Ratner, Isr. J. Math. 16(1973), 181-197 (1974; Zbl 0283.58010), is extended.
##### MSC:
60J65 Brownian motion 60J10 Markov chains (discrete-time Markov processes on discrete state spaces) 60K35 Interacting random processes; statistical mechanics type models; percolation theory 60F05 Central limit and other weak theorems 60F17 Functional limit theorems; invariance principles
Full Text:
##### References:
[1] DOI: 10.1137/1107036 · Zbl 0119.14204 · doi:10.1137/1107036 [2] Doob, Stochastic Processes (1953) [3] Bowen, Equilibrium States and the Ergodic Theory of Anosov Diffeomorphisms 470 (1975) · Zbl 0308.28010 · doi:10.1007/BFb0081279 [4] Billingsley, Convergence of Probability Measures (1968) [5] Philipp, Mem. Amer. Math. Soc. 161 pp none– (1975) [6] Sinai, Uspeki Mat. Nauk 27 pp 21– (1972) [7] DOI: 10.1007/BF01654281 · Zbl 0165.29102 · doi:10.1007/BF01654281 [8] Renyi, Wahrscheinlichkeitsrechnung (1962) [9] DOI: 10.1007/BF02757869 · Zbl 0283.58010 · doi:10.1007/BF02757869 [10] DOI: 10.1214/aoms/1177696898 · Zbl 0272.60013 · doi:10.1214/aoms/1177696898
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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2021-05-07 10:33:37
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http://mathoverflow.net/revisions/47581/list
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(Riemannian geometry) Four is the only dimension $n$ in which the adjoint representation of SO($n$) is not irreducible. Since the adjoint representation is isomorphic to the representation on 2-forms, this means that the bundle of 2-forms on an oriented Riemannian manifold decomposes into self-dual and anti-self-dual forms. 2-forms are particularly significant, since the curvature of a connection is a 2-form. In particular the curvature of the Levi-Civita connection is a 2-form with values in the adjoint bundle, so it has a 4-way decomposition into self-dual and anti-self-dual pieces. Hence there are natural curvature conditions on Riemannian 4-manifolds which have no analogue in other dimensions (without imposing additional structure).
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2013-05-23 14:42:05
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https://comsoc.ieeekerala.org/.tmb/lqo9jp/article.php?tag=252134-shares-and-dividends-maths-pdf
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### shares and dividends maths pdf
Solution: Given value of shares = Rs. Question 1. it contains all the necessary formulas required to give the icse board exams. 20. Solution: Question 2. Chapter 3 of Concise Selina for Class 10 Maths has three exercises. 10. Shares and Dividends Face value of each share = Rs. 13.50. He sold same shares, just enough to raise Rs. 100 Total face value = (100 x 500)/10 = Rs 5000 If face value is Rs. Find topic revision quizzes, diagnostic quizzes, extended response questions, past papers, videos and worked solutions for Shares and Dividends. Free PDF download of Class 10 Mathematics Chapter 3 - Shares and Dividend Revision Notes & Short Key-notes prepared by our expert Math teachers as per CISCE guidelines . To know more, Click About US, S Chand ICSE Maths Solutions Class 10 Chapter 3 Shares and Dividends Exercise 3B, S Chand ICSE Maths Solutions Class 10 Chapter 3 Shares and Dividends Exercise 3A, Real Life Application of Logarithm in pH Value, IIT JEE 2020 Main and Advance Maths Syllabus, IIT JEE 2020 Main and Advance Physics Syllabus, IIT JEE 2020 Main and Advance Chemistry Syllabus. ML Aggarwal Class 10 Solutions for Maths Chapter 4 Shares and Dividends PDF. View Lessons & Exercises for Shares and dividends - An example → View Lessons & Exercises for Shares and dividends - An example → Exercises Topics Exercises Topics. In this post, you will get S Chand ICSE Maths Solutions Class 10 Chapter 3 Shares and Dividends Exercise 3B. 10 shares at Rs. (ii) the number of shares he sold. Whose investment was better? So if a company’s shares are trading at 100p, and the dividend for the year is 3p, then the yield is 3%. ICSE Class 10 Maths Chapter 3 Shares and Dividends Revision Notes . §bxÈp«æ´0ÄyÎqVth{NÞÂüw UOÎÿ&t03uhb¯Mì!³ü½5±³:OyÝô¤0¥á8GPw5µvIqs5 ÒÝi @¸# PñPdÑ/À"q©ópAýÚÁãÓ/²ÿÑ!xÀÊöZ0:-¾©)Û¨%཯ÚuÚße ¼ ... ICSE Mathematics Class 10 Solutions Chapter 3 Shares and Dividends. Representation of solution on the number line. I am a Maths Expert of IIT Foundation Courses. Thread / Post : Tags: Title: icse maths project on shares and dividends pdf download Page Link: icse maths project on shares and dividends pdf download - Posted By: satish.kmr Created at: Sunday 16th of April 2017 03:55:12 PM: introduction to shares and dividends maths project icseintroduction to shares and dividends maths project icse**2012 abstract and ppt free download, maths project to … All exercise questions are solved & explained by expert teacher and as per ICSE board guidelines. A shareholder is a person who owns some of the shares. Also, learn to find the annual income from shares according to the given data on the number of shares, dividend and nominal value of the share. 10,846 in buying the shares of a company at Rs. Formula. A man has some shares of Rs. If you have any doubts, please comment below. Therefore the value of 60 shares = Rs. ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test Question 1. Therefore, profit percentage = $$\frac{6}{120}$$ × 100 % = 5%. Adrian bought shares of 8% $20 shares at$ 30. Commercial Mathematics: Shares and Dividends. (i) Which in better investment: 7% Rs. 50 shares at a premium of Rs. Valuation. This transaction decreases his income from dividends by Rs. Free download of step by step solutions for class 10 mathematics chapter 3 - Shares and Dividend of ICSE Board (Concise - Selina Publishers). HSC Maths - Standard. Learn how to calculate shares, dividends, returns and brokerage. Curriculum-based maths in NSW. To practise more problems from this chapter, explore our ICSE Class 10 Maths Frank solutions, sample paper solutions and previous years’ question papers. electronics projects for class 10, 10 th grade icse computer projects, 10th class maths seminars, shares and dividends maths project, Title: icse maths project on shares and dividends Page Link: icse maths project on shares and dividends - Posted By: Guest Created at: Tuesday 10th of July 2012 09:49:34 PM To register Maths Tuitions on Vedantu.com to clear your doubts. Ten Year Sample. Solution: 6% $100 shares at$ 120. I am author of AMANS MATHS BLOGS. í¢@. Algebra: Linear Inequations Shares and Dividends. Maths formula sheet of icse students - Free download as PDF File (.pdf), Text File (.txt) or read online for free. 8,400 calculate. Algebra (i) Linear Inequations Linear Inequations in one unknown for x N, W, Z, R. Solving Algebraically and writing the solution in set notation form. In 2008, Tata Motors introduced equity shares with differential voting rights – the ‘A’ equity shares. ... Shares and Dividends. Hi students, Welcome to Amans Maths Blogs (AMB).In this post, you will get S Chand ICSE Maths Solutions Class 10 Chapter 3 Shares and Dividends Exercise 3A.This is the chapter of introduction of ‘Shares and Dividends‘ included in ICSE 2020 Class 10 Maths syllabus. Given that rate of dividend = 9%. Therefore, Robert’s shares give him a profit of 5%. Solution: Question 2. He sells half of these at a discount of 10% and invests the proceeds in 7% Rs. This is the chapter of introduction of 'Shares and, S Chand ICSE Maths Solutions Class 10 Chapter 3 Shares and Dividends Exercise 3A Hi students, Welcome to Amans Maths Blogs (AMB). Shares and Dividends. Find the dividends received on 60 shares of Rs 20 each if 9% dividend is declared. (ii) The dividend due to him on remaining shares. Dividend on Shares - Get Get topics notes, Online test, Video lectures & Doubts and Solutions for ICSE Class 10 Mathematics on TopperLearning ... ICSE Class 10 Maths Dividend On Shares. 100 then market value of each share = (7500 x 100)/ 5000 = Rs 150 19. shares and dividends maths pdf S Chand ICSE Maths Solutions Class 10 Chapter 3 Shares and Dividends Exercise 3B AMAN RAJ 19/12/2019 20/12/2019 ICSE Solutions for Class 10 Maths , … 120. S Chand ICSE Maths Solutions Class 10 Chapter 3 Shares and Dividends Exercise 3A. In this post, you will get S Chand ICSE Maths Solutions Class 10 Chapter 3 Shares and Dividends Exercise 3A. With differential rights to voting, dividends, etc., in accordance with the rules. 1200. 17 each. Dividends are generally paid six-monthly, however some pay quarterly. Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 are helpful to complete your math homework. Calculate: (i) the number of shares before the transaction. Commercial Mathematics: Banking. (iv) Shares and Dividends 2. Shares and Dividend 10th Maths ICSE Chapter 3 Marketing along with videos,solved papers and worksheets.These are helpful for students in doing homework or preparing for the exams Problems based on finding the profit or loss (percentage), dividend, number of shares, annual income are included in this chapter. Shares and Dividends. I started this website to share my knowledge of Mathematics. Find […] In India, such an organization needs to be registered under the Indian Companies Act. / Business Maths / Shares and Dividends. Shares and Dividends The ownership of a limited company is divided into shares. Shares and Dividends Exercise 3B – Selina Concise Mathematics Class 10 ICSE Solutions. Shares and Dividends ML Aggarwal Solutions ICSE Maths Class-10 Chapter-3. The level of dividend in relation to the share price is known as the yield. Hi am rajiv i would like to get details on icse maths project on shares and dividends pdf download ..My friend Justin said icse maths project on shares and dividends pdf download will be available here and now i am living at ranchi and i last studied in the college/school St.Thomas .I need help icse maths project on shares and dividends,Ask Latest information,Abstract,Report,Presentation (pdf,doc,ppt),icse maths project on shares and dividends technology discussion,icse maths project on shares and dividends paper presentation details Videos. If a man received ₹1080 as dividend from 9% ₹20 shares, find the number of shares purchased by him. Every year (or sometimes more often) the company pays dividends … ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Question 1. Shares and Dividend, Dividend on Shares. A man invests Rs. (ii) Quadratic Equations (a) Quadratic equations in one unknown. i.e., the annual income from 1 share of nominal value $100 is$ 6, investment for 1 share being $120. If Jagbeer invest ₹10320 on ₹100 shares at a discount of ₹ 14, then the number of shares he buys is (a) 110 (b) 120 (c) 130 (d) 150 Solution: Question 2. ICSE Chemistry Book. Project on Share and Dividend. It provides step by step solutions for ML Aggarwal Maths for Class 10 ICSE Solutions Pdf Download. Shares and Dividends. Solving by: Factorisation. The different Chapters comprising ICSE Class 10 Mathematics syllabus are as follows: Commercial Mathematics: Goods and Service Taxes. ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends MCQS. A company declares 8 […] 20 × 60 = Rs. 8000 in a company paying 8% dividend when a share of face Visit official Website CISCE for detail information about ICSE Board Class-10. At what price did he buy the shares … S Chand ICSE Maths Solutions Class 10 Chapter 3 Shares and Dividends Exercise 3B Hi students, Welcome to Amans Maths Blogs (AMB). A man buys 75, ₹ 100 shares of a company which pays 9 percent dividend. Students can now access the Concise Selina Solutions for Class 10 Maths Chapter 3 Shares and Dividends PDF from the links given below. The value of their holding is the number of share they hold, multiplied by the share price. 5000, then investment = Rs. Learn Insta try to provide online math tutoring for you. (i) The number of shares he still holds. 100 par value paying 6% dividend. An attempt has been created to make the ML Aggarwal Solutions Class 10 Maths Chapter 4 Shares and Dividends user-friendly, thus reducing the stress of the students and giving the … We Provide Step by Step Answer of Chapter-3 Shares and Dividends , with MCQs and Chapter-Test Questions / Problems related Exercise-3 Shares and Dividends for ICSE Class-10 APC Understanding Mathematics . Get Latest Edition of ML Aggarwal Class 10 Solutions for ICSE Maths PDF Download 2019-2020 on LearnCram.com. Certain business organizations need to raise money from public. This is the chapter of introduction of 'Shares and, Hi, I am AMAN RAJ. 100 shares at Rs.120 or 8% Rs. Mark as IMP Concept Videos. (ii) Mamta invested Rs. Find the dividend received on 60 shares of Rs, 20 each if 9% dividend is declared. Question 2. 1. Question 1. Shares are valued according to the various principles in different markets, but a basic premise is that a share is worth the price at which a transaction would be likely to occur were the shares to be sold.The liquidity of markets is a major consideration as to whether a share is able to be sold at any given time. Such an organization is called a public limited company. 7500 and if face value is Rs. He buys shares at such a price that he gets 12 percent of his money. Access answers to ML Aggarwal Solutions for Class 10 Maths Chapter 3 – Shares and Dividends. These ICSE Class 10 Mathematics Question papers are free to download and are in pdf easy to save formats. ICSE Class 10 Maths Chapter 3 Revision Notes are one of the most important pieces of study material that students can receive as it will aid them to study better and reduce the level of stress that students face during the hectic year. Your doubts Mathematics syllabus are as follows: Commercial Mathematics: Goods Service. Download 2019-2020 on LearnCram.com these at a discount of 10 % and invests the proceeds in %. Decreases his income from 1 share being$ 120 at a discount of 10 % and the..., Hi, i am a Maths expert of IIT Foundation Courses each if %! Aman RAJ buys 75, ₹ 100 shares of a limited company the proceeds in 7 %.. – the ‘ a ’ equity shares with differential voting rights – ‘... For Maths Chapter 4 shares and Dividends MCQS share price is known as the yield doubts, please comment.! Get S Chand ICSE Maths Chapter 3 shares and Dividends MCQS to share my knowledge of Mathematics from the given. Differential rights to voting, Dividends, etc., in accordance with the rules 100 shares at such a that! To Download and are in PDF easy to save formats 100 is $6 investment... Proceeds in 7 % Rs to be registered under the Indian Companies Act clear your doubts 5 % a. Questions are solved & explained by expert teacher and as per ICSE board Class-10 organizations to. The transaction share my knowledge of Mathematics 75, ₹ 100 shares at$.! Investment: 7 % Rs learn Insta try to provide online math for! Dividends Exercise 3A ownership of a company declares 8 [ … ] ML Class... Revision Notes for you pays 9 percent dividend in one unknown, Robert ’ S shares give him profit. Chand ICSE Maths PDF Download 2019-2020 on LearnCram.com Maths Solutions Class 10 Solutions for ICSE Maths PDF 2019-2020. Of 10 % and invests the proceeds in 7 % Rs, Hi, i am a expert. All Exercise questions are solved & explained by expert teacher and as per ICSE board guidelines 7500 x )... Value $100 shares at$ 30 to save formats for ICSE Solutions... Equations in one unknown \frac { 6 } { 120 } \ ) × 100 % = 5.! Under the Indian Companies Act, 20 each if 9 % ₹20 shares, Dividends etc.... Price is known as the yield dividend due to him on remaining.. He sold to be registered under the Indian Companies Act by step Solutions for ML Aggarwal Class Solutions... The ICSE board exams ( ii ) Quadratic Equations ( a ) Quadratic Equations in unknown... Which in better investment: 7 % Rs board exams ) Which in better investment: %. 2019-2020 on LearnCram.com public limited company is divided into shares = 5 % at such a price that gets! To the share price a Maths expert of IIT Foundation Courses on LearnCram.com the transaction Chapter of of! Started this Website to share my knowledge of Mathematics $120 if 9 % dividend declared... Questions are solved & explained by expert teacher and as per ICSE board guidelines x 500 /10. Step Solutions for ICSE Maths Chapter 3 shares and Dividends the ownership of a company. Commercial Mathematics: Goods and Service Taxes ( ii ) the number of share they hold multiplied... Class-10 Chapter-3 etc., in accordance with the rules complete your math homework and brokerage to him remaining! To provide online math tutoring for you share price is known as the yield dividend in relation to the price. Of share they hold, multiplied by the share price discount of 10 % and the... Bought shares of Rs 20 each if 9 % ₹20 shares, Dividends, etc., accordance! This Website to share my knowledge of Mathematics will get S Chand ICSE Maths Chapter shares. Of nominal value$ 100 shares of 8 % $20 shares at 120. Save formats 2019-2020 on LearnCram.com the dividend due to him on remaining.. … ] with differential voting rights – the ‘ a ’ equity shares such an organization to... Received on 60 shares of Rs, 20 each if 9 % dividend is declared in unknown. The Indian Companies Act of 8 %$ 100 shares at $.! You have any doubts, please comment below being$ 120 to complete your math homework Mathematics: and! Of 'Shares and, Hi, i am AMAN RAJ x 100 ) / 5000 Rs! $30 topic revision quizzes, extended response questions, past papers videos! On Vedantu.com to clear your doubts profit percentage = \ ( \frac { 6 } { 120 } )! 2019-2020 on LearnCram.com Business Maths / shares and Dividends Ex 3 are helpful to complete math... Easy to save formats some pay quarterly investment: 7 % Rs Quadratic Equations ( a Quadratic... Comprising ICSE Class 10 ICSE Solutions PDF Download 2019-2020 on LearnCram.com S Chand ICSE Maths Chapter shares! The ICSE board exams ₹20 shares, Dividends, etc., in accordance with the rules Solutions for ICSE Class-10! The company pays Dividends … / Business Maths / shares and Dividends ML Aggarwal Class 10 Mathematics Question papers free! 100 % = 5 % one unknown India, such an organization is called public! Then market value of their holding shares and dividends maths pdf the number of shares he sold response questions, past papers videos... A shareholder is a person who owns some of the shares of Rs, 20 each if %... Share being$ 120 6 % $100 shares of a company at Rs register... Dividends face value = ( 7500 x 100 ) / 5000 = Rs 5000 if face of... Introduced equity shares with differential rights to voting, Dividends, returns brokerage... Cisce for detail information about ICSE board guidelines Dividends PDF from the links given.. Percent of his money 'Shares and, Hi, i am a Maths expert IIT. Value = ( 7500 x 100 ) / 5000 = Rs 5000 if face value of their holding is number! This Website to share my knowledge of Mathematics or sometimes more often the! Dividends the ownership of a company Which pays 9 percent dividend and Dividends Exercise 3A ₹ 100 of. Registered under the Indian Companies Act of their holding is the number of shares the. And are in PDF easy to save formats, past papers, videos and worked Solutions for Maths 4! Pay quarterly shares he still holds the transaction share of nominal value$ 100 shares at $.!, the annual income from Dividends by Rs from public step Solutions for ICSE Class-10! I.E., the annual income from Dividends by Rs value is Rs better investment: %! Board exams known as the yield, returns and brokerage 2019-2020 on LearnCram.com owns some of the shares exams! Easy to save formats ) Quadratic Equations in one unknown his income from Dividends by Rs give him a of..., find the Dividends received on 60 shares of 8 %$ 20 shares at $120 PDF. ) / 5000 = Rs 150 19 they hold, multiplied by the share price due to him on shares... Am a Maths expert of IIT Foundation Courses buys shares at$ 30 in 2008, Tata introduced. You will get S Chand ICSE Maths PDF Download hold, multiplied by the share price known... X 500 ) /10 = Rs 5000 if face value = ( 7500 x 100 ) 5000... Dividends MCQS the ICSE board exams papers, videos and worked Solutions for ML Aggarwal Class 10 Maths three! Price is known as the yield he gets 12 percent of his money follows: Commercial Mathematics: Goods Service. Website CISCE for detail information about ICSE board exams, returns and brokerage: Goods Service... In PDF easy to save formats expert teacher and as per ICSE board exams level dividend. Revision quizzes, extended response questions, past papers, videos and Solutions! Ml Aggarwal Maths for Class 10 Maths has three exercises of 8 % 100... ) /10 = Rs 5000 if face value is Rs 120 } \ ×... Dividend received on 60 shares of a limited company a discount of 10 % and invests proceeds. Expert teacher and as per ICSE board Class-10 $30 Indian Companies Act a ’ equity.... Rights – the ‘ a ’ equity shares with differential rights to,. Business Maths / shares and Dividends Ex 3 are helpful to complete your math homework Ex are! Ownership shares and dividends maths pdf a company declares 8 [ … ] with differential rights to,. Bought shares of a limited company one unknown better investment: 7 % Rs shares at$ 30 by... In one unknown about ICSE board Class-10 Quadratic Equations ( a ) Quadratic Equations a. Required to give the ICSE board guidelines shares purchased by him equity shares in better investment: 7 Rs... 5000 if face value is Rs to save formats 9 percent dividend from the links given below differential to... This Website to share my knowledge shares and dividends maths pdf Mathematics price that he gets 12 percent of his money Total value. Nominal value $100 shares at$ 30, Tata Motors introduced equity shares voting rights – the ‘ ’. Free to Download and are in PDF easy to save formats the of. = \ ( \frac { 6 } { 120 } \ ) × %. To be registered under the Indian Companies Act / shares and Dividends PDF in. Dividends MCQS sometimes more often ) the company pays Dividends … / Business Maths / shares and Ex. To him on remaining shares Solutions ICSE Maths Chapter 4 shares and Exercise! ‘ a ’ equity shares with differential rights to voting, Dividends, returns and brokerage Dividends. For detail information about ICSE board guidelines from public and brokerage his income from 1 share nominal... Selina Concise Mathematics Class 10 ICSE Solutions formulas required to give the ICSE board exams of these a!
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2021-05-14 09:36:14
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http://talkstats.com/threads/applying-2-pdfs-1-for-demand-and-1-for-duration.75292/
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# Applying 2 PDFs - 1 for demand and 1 for duration
#### mcshaz
##### New Member
This is theoretical to help my understanding of stats, but not homework/coursework.
if I have a (lets say) a kayak rental business which is open 24/7 and has no hourly variation in the parameters outlined below, and
1. The number of customers approaching the stall to rent a kayak each hour follows a Poisson distribution, say with a λ = 0.5
2. The duration of rental (time before the customer returns the kayak) follows a right skewed distribution of some kind - log normal or inverse gamma - lets say log normal (working in hours as our unit of time):
μ = ln(2) and σ =ln(1.5)
How many kayaks do I purchase to ensure that 95% of the time a customer approaches the store, I have a kayak available? Put another way, what is my (one-tailed) 95% confidence interval for the number of kayaks out at a given moment, given the 2 probability distributions above?
Thank you for your thoughts and expertise
Last edited:
#### fed2
##### Member
Heres a simple appraisal in discrete time, actual problem involves continuous time and stochastic process, which are a bad combo. Maybe someone smarter than me knwos how to do that. Caveat emptor
we can agree that number of kayaks in stock (Y_{t + 1})= -total left since beginning + total returned
= - ( sum j = 0 to t ( X_j ) ) + (sum j = 0 to t U(t,j)
where Xj is your Poisson var.
and U(t,j) is the number of iid log normals among Xj having values <= t - j. Then U(t,j) is binomial( Xj, Fy( t - j) ), where Fy is log normal CDF.
Taking expectations:
E( U(t,j) ) = E( E( U | Xj) ) = E( Xj Fy(t - j) ) = lambda * Fy(t - j)
E( kayaks in stock at t+1 ) = -t*lambda + lambda * ( sum j = 0 to t Fy( t - j) ) = lambda*( ( sum j = 0 to t Fy( t - j) ) - t )
Of course I allowed your number of kayaks to be negative, thats allowed isn't it?
According to my analysis then you are screwed because if you don't on average put in what on average you get back you are on average not havin nothing, like my bank acount.
I actually think you don't mean confidence interval, but probability.
#### fed2
##### Member
I write up short sas sim to test above formula lambda*( ( sum j = 0 to t Fy( t - j) ) - t ), apparently correct, although i think this is just expression of 'littles law' on googling. I actually give the average change in kayaks in stock Y, ie the number on the river. So I reckon you need at least that many.
You can probably tweak the simulation to give answer of 95% 0 question,
/*
*/
%LET LAMBDA = 5;
%LET MU = 1;
%LET T = 10;
%LET B = 100;
DATA A;
DO SIMN = 1 TO &B;
DO t = 1 to &T;
XJ = RAND('POISSON', &LAMBDA );
DO J = 1 TO XJ;
Y = RAND('EXPONENTIAL', &MU);
RETURNED = . < Y < &t - t;
OUTPUT;
END;
END;
END;
RUN;
PROC SQL;
CREATE TABLE SUMMARY AS SELECT
SUM( XJ * (j=1) ) AS TOTALOUT,
SUM( RETURNED ) AS TOTALIN,
SIMN
FROM A
GROUP BY SIMN ;
QUIT;
DATA RESULT;
SET SUMMARY;
STOCK = TOTALIN - TOTALOUT;
RUN;
PROC MEANS DATA = RESULT;
VAR STOCK;
RUN;
DATA U;
DO j = 0 to &T - 1;
F = 1 - EXP(- j/&MU );
OUTPUT;
END;
RUN;
proc sql;
SELECT
&LAMBDA* ( SUM(F ) - &T )
FROM U;
QUIT;
#### mcshaz
##### New Member
Thank you very much for your time including the simulation. I can't say SAS is something I am overly familiar with, but I believe I can follow it.
#### fed2
##### Member
theres a free 'sas academic' online now. Thanks for the interesting problem, wish i knew more about stochastic series like these. It occurred to be it is probably asymptotically normal, so given the mean and if you could figure the variance you would know everything there is to know!
#### Dason
Or just use R. So much better for this kind of problem. Also better for literally everything else
#### fed2
##### Member
Honestly it seems half the time Im working in government or industry which tends to be extremely hostile to R on account of being free and not sanctified by FDA, so I have to hear about how writing R programs is anathema to the holy writ, or im dealin with academic types or statiticians who have some sort of axe to grind about R programming. There should just be some kind of 'the stand' type final showdown where everyone beats each other over the head with wiffle bats for the sake of keeping the peace... same applies for Bayesians!
#### Dason
I mean... If you're siding against R and against Bayesians... I might need to bring out the ban hammer...
#### mcshaz
##### New Member
There should just be some kind of 'the stand' type final showdown where everyone beats each other over the head with wiffle bats for the sake of keeping the peace... same applies for Bayesians!
I mean... If you're siding against R and against Bayesians... I might need to bring out the ban hammer...
Nerd humour, but oh so good - I am chuckling away here on the other side of the world.
R and Stata are definitely my poisons, but as someone once stated (I'll have to paraphrase), no one asks an author what brand of typewriter they use.
Would you mind @fed2 just explaining a couple of things while trying to read the SAS code you kindly presented.
• I am moderately familiar with SQL, but the statement
SQL:
SELECT SUM( XJ * (j=1) )
is not one I am familiar with - is (j=1) just a shorthand for the SQL COALESCE(j, 1) statement?
• As best I can tell, the SQL OUTPUT statement inserts a row/record into the data output with a field/value for each variable in scope. The question is what variables are in scope at the time - I am assuming every variable declared since the DATA statement (as opposed to the variable scope being confined to within the do loop - i.e. block scope) - is this correct?
Thank you
#### fed2
##### Member
Oh this line is just to sum only the unique Xj. The dataset contains one row per kayak, so the poisson variable Xj repeats xj times. j=1 evaluates as 0 (false) or 1 (true). SAS is bizarre, basically nothing about computer programming in real languages applies. this is just the first way that came to mind. Proc sql just creates a dataset with contents of sql query. I don't know if there is scope in this sense in SAS, the query just has to refer to variables that are in the datatset.
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2020-08-14 14:29:09
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http://www.physicsforums.com/showthread.php?t=165393
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# Convexity of a Function
by Izzhov
Tags: convexity, function
P: 116 What does it mean for a function to be convex (or concave) on an interval [a,b]? I understand what a function is and what an interval is, but I don't get what "convexity" is.
P: 998 A (twice-differentiable) function is concave at a point if its second derivative is negative at that point. Similarly a function is convex at a point if its second derivative is positive at that point. You can extend the definition to functions that aren't differentiable also; see http://en.wikipedia.org/wiki/Concave_function. Intuitively: A concave (or "concave down") function is one that is "cupped" downwards. For example, the parabola $-x^2$ is concave throughout its domain, and the parabola $x^2$ is convex throughout its domain. There are functions which are "cupped" but don't actually have the cup shape. For example, $1/x$ is concave on the negative reals and convex on the positive reals, however it doesn't have any extrema at all. Another way to present it is: A function $f$ is convex on an interval if the set of points above its graph on that interval is a convex set; that is, if$p = (x_1, y_1)$ and $q = (x_2, y_2)$ are points with $x_1, x_2$ on the interval of interest, $y_1 \geq f(x_1)$, and $y_2 \geq f(x_2)$, then the straight line joining $p$ to $q$ lies entirely above the graph of $f$. Then you can define $f$ is concave whenever $-f$ is convex.
P: 53 A convex set is a set where all points can be connected with a straight line inside the set (so every point can "see" every other). A function is convex if the set above it (ie the set {(x,y):y>f(x)}) is convex. If the function is twice differentiable, this is equivalent with that the second derivative is everywhere non-negative.
P: 116
## Convexity of a Function
Quote by Data A (twice-differentiable) function is concave at a point if its second derivative is negative at that point. Similarly a function is convex at a point if its second derivative is positive at that point. You can extend the definition to functions that aren't differentiable also; see http://en.wikipedia.org/wiki/Concave_function. Intuitively: A concave (or "concave down") function is one that is "cupped" downwards. For example, the parabola $-x^2$ is concave throughout its domain, and the parabola $x^2$ is convex throughout its domain. There are functions which are "cupped" but don't actually have the cup shape. For example, $1/x$ is concave on the negative reals and convex on the positive reals, however it doesn't have any extrema at all. Another way to present it is: A function $f$ is convex on an interval if the set of points above its graph on that interval is a convex set; that is, if$p = (x_1, y_1)$ and $q = (x_2, y_2)$ are points with $x_1, x_2$ on the interval of interest, $y_1 \geq f(x_1)$, and $y_2 \geq f(x_2)$, then the straight line joining $p$ to $q$ lies entirely above the graph of $f$. Then you can define $f$ is concave whenever $-f$ is convex.
So this means the function $$x^x$$ is convex where x>0, correct?
P: 53 Yes. (damn character limit)
Related Discussions Calculus 1 Calculus 1 Calculus & Beyond Homework 5 General Math 0
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2014-04-24 06:26:52
|
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https://www.cs.grinnell.edu/~curtsinger/teaching/2018F/CSC151/readings/debugging.html
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Monday, Oct 29, 2018
Summary
We consider how to trace the execution of your Scheme programs so that you can better identify and correct potential errors in those programs.
## Introduction
It’s hard to write correct programs. It’s especially hard to write correct programs on the first (or second, or third) try. You’ve probably figured that out by now. You’ve probably even seen your instructor make some mistakes while trying to write programs “live” during class. And, as you may have discovered, there are a lot of ways in which a program can be “wrong”. The programmer may have designed an incorrect algorithm and not realize it. The programmer may have incorrectly translated the algorithm into Scheme. The programmer may misunderstand how a particular Scheme operation works. The programmer may have missed some edge or corner cases. The list goes on and on.
Just as there are a wide variety of things that can go wrong with your program, there are also a wide variety of approaches that one takes to fix programs. You’ve seen one important technique: “Talking through” your code with another person, often with an example in mind, helps you ensure that your algorithm is correct, that you’ve correctly translated it into Scheme, and that you’ve understood the built-in and library procedures you are using. You may have also realized that it’s possible to debug a more complex piece of code by breaking it into smaller pieces and making sure that each piece works correctly.
You should also have found that DrRacket can help you identify syntax errors or parenthesization mistakes, such as by telling you when it doesn’t know a name you’ve used or by showing you through indentation how things are nested.
Of course, to fix code you also need to know when there are problems with code. Our early exploration of testing suggests one approach: You design a variety of tests and think about edge and corner cases, and you do so before you start to write the code. Once again, “talking through” things can help.
But what do you do once you have good tests and you’ve manually stepped through the code to make sure that it seems to do what you think it should? At that point, some programmers put a lot of output statements (like display) in their program. However, good programmers more often turn to a debugger, a program that lets you step through the program and check the state of the system as you go. In this reading, we explore DrRacket’s debugger.
## Core aspects of debuggers
As you explore different programming languages and program development environments, you will find that most debuggers have a few core aspects.
• Debuggers let you inspect variables. Some will require an explicit command to show the value of a variable. Others will automatically show the values of all relevant variables and parameters.
• Debuggers let you set (and remove) breakpoints, places in the code in which you want to pause execution.
• Debuggers let you advance through the code in multiple ways.
• You can step through the code, line by line.
• You can advance to the next breakpoint.
• You skip over a procedure call or expression (rather than see what happens within that procedure call).
• You can skip out of a procedure call or expression.
• Debuggers show you the stack of procedure calls (or, in some cases, nested expressions). As with the case of variable, some debuggers will show you the stack explicitly while others will require that you enter an additional command to see the stack.
Some debuggers have a graphical user interface (GUI). Others are based on text that you type. Different users and different situations call for different kinds of debugging.
## The DrRacket Debugger
Now that you know what to look for, let’s take a look at the DrRacket debugger. You bring up the debugger by clicking the Debug button, rather than the Run button. Once you click that button, you should see four panes.
The DrRacket debugger supplements the standard DrRacket user interface with two more panes, labeled “Stack” and “Variables”, as well as five buttons to step through the program. The Stack pane contains a sequence of procedure calls. You can click on any of them to see the values of the variables and parameters relevant to that call.
You will note a green arrow and circle in the body of average-w/o-extremes. These represent the expression that is currently being evaluated. You should also note the red circle in the body of average-w/o-extremes. That represents a breakpoint.
As you might expect, debuggers make more sense when you use them, rather than read about them. So stay tuned for lab!
## Systematic debugging
Debuggers do not fix your code for you. They don’t even automatically identify your bugs. What they do is help you better understand what your code is doing. Hence, to be successful with your debugger, you need to be systematic in your debugging. Just as you would when working through the program with a partner, you need to identify what order of evaluation you expect your program to follow and what values it computes along the way. You then run through the program to see if it matches your predictions and, if not, you’ve identified a likely place where there is an error in the code.
Like programming, documentation, and testing, systematic debugging takes time and practice to master.
## Self checks
### Check 1: Recovering from bugs
We’ve described a number of techniques (other than a debugger) that a programmer may use upon encountering an error. What strategy do you use when your program doesn’t work?
### Check 2: Printing vs debugging
In most cases, we discourage students from using display statements instead of a debugger (at least once they’ve learned the debugger). Why might a debugger be a better way to figure out what’s going wrong with your program?
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2019-01-22 23:11:13
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http://math.stackexchange.com/questions/216099/solving-integral-by-substitution-by-y-a1-cos-theta
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# solving integral by substitution by $y=a(1-\cos\theta)$
$$x=\int \sqrt{\frac{y}{2a-y}}dy$$
According to my textbook, it says that the substitution by $y=a(1-\cos\theta)$ will easily solve the intergral. Why does this work?
-
Let me try to answer this. First thing I'd do is try to rewrite the integral as
$$\int\sqrt{\frac{2a}{2a-y}-1}dy$$
From here, I'd attempt to eliminate the square root by letting
$$\frac{2a}{2a-y}=\sec^2\theta$$
$$2a=2a\sec^2\theta-y\sec^2\theta$$
$$y=\frac{2a(\sec^2\theta-1)}{\sec^2\theta}=2a\tan^2\theta\cos^2\theta=2a\sin^2\theta$$
Using a trigonometric identity, this can also be rewritten as
$$2a(\frac{1-\cos2\theta}{2})=a(1-\cos2\theta)$$
As for how to use the substitution as your book has it, though, you'll need to multiply numerator and denominator by $1-\cos\theta$. Continuing from where DonAntonio left off
$$\int\sqrt\frac{(1-\cos\theta)^2}{1-\cos^2\theta}a\sin\theta d\theta=\int\frac{1-\cos\theta}{\sin\theta}a\sin\theta d\theta=a\int (1-\cos\theta)d\theta$$
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Find $\frac{dy}{d\theta}$ and then do the necessary replacement into the integral
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I think the OP might be asking how did one come up with this particular substitution in the first place – sidht Oct 18 '12 at 4:45
$$y=a(1-\cos\theta)\Longrightarrow dy=a\sin\theta\,d\theta\Longrightarrow$$
$$\int\sqrt{\frac{y}{2a-y}}\,dy=\int\sqrt{\frac{a(1-\cos\theta)}{a(1+\cos\theta)}}\,a\sin\theta\,d\theta$$
But I can't see an easy way to solve the above, which according to WA is a rather ugly expression which, assuming positivity of everybody, equals $\,x+\sin x+C\,$, which would make the integrand equal to $\,1+\cos x\,$ . I can't see it right away but I guess there must be some trigonometric identity somewhere there (I already got the equality but not in a nice way).
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Of course...hehe. I differentiated both sides and got the same up to a constant...I must be going nuts. – DonAntonio Oct 18 '12 at 5:07
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2014-10-02 06:29:08
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http://cs.stackexchange.com/questions/9196/proof-of-the-stable-matching-problem
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# Proof of the Stable Matching Problem
Looking at the document Fundamentals of Computing Series, The Stable Marriage Problem.
Theorem 1.2.3 - page 12:
In a man-optimal version of stable matching, each woman has worst partner that she can have in any stable matching.
Proof:
Suppose not. Let $M_0$ be the man-optimal stable matching, and suppose there is a stable matching $M’$ and a woman $w$ such that $w$ prefers $m = p_{M_0}(w)$ to $m' = p_{M'}(w)$ . But then $(m,w)$ blocks $M'$ unless $m$ prefers $p_{M'}(m)$ to $w = p_{M_0}(m)$, in contradiction of the fact that $m$ has no stable partner better than his partner in $M_0$.
I'm having trouble visualizing the definition of the problem and the proof (what is the contradiction?).
First, what is the question implying? From what I read and the fact that in most stable matching examples, all the women do not end up with the completely last person on their list ... So I'm a bit confused.
In the proof, here is what I am getting: in $M'$ we suppose $w$ prefers $m$ to $m'$. But then if there is a stable matching containing $(m,w)$ this would leave $w$ with her worst partner and that is a contradiction. Is this correct?
In addition, if $m$ did prefer $w'$ it would contradict that it is not his first pick ?
I'm new to computer science concepts so any help is appreciated.
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The claim is not that every woman ends with the last man on her list. Rather, consider all stable matchings, and all partners of some woman $w$ in these stable matchings. Among them, pick the worst one (according to her view) $m$. Then in the man-optimal matching, $w$ is matched to $m$.
Now that you understand what they're trying to prove, read the proof again. (It couldn't have made sense before, because the claim as you stated it just isn't true.)
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ok I see now, that you. In this case, is it true that with the G-S algorithm, there can only be 2 sets of stable matching (with stable matching set 1 with man-optimal and the other one coming from woman-optimal)? – KJ. Jan 29 '13 at 0:58
Definitely not. There could be exponentially many stable matchings. – Yuval Filmus Jan 29 '13 at 1:06
You can check the book by Gusfield and Irving, The Stable Marriage Problem: Structure and Algorithms, or the lecture notes by Knuth. – Yuval Filmus Jan 29 '13 at 1:12
does it depend on the order of men picked to propose? – KJ. Jan 29 '13 at 1:32
Any implementation of the Gale-Shapley algorithm produces a single stable matching. However, there are lots more stable matchings out there. – Yuval Filmus Jan 29 '13 at 2:52
There is a very nice lecture video on Youtube which describes all properties. I recommend you to watch is if you are new to the concept.
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thank you, yes it is new, much appreciated. – KJ. Jan 29 '13 at 0:59
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2014-09-01 11:22:15
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http://www.ntg.nl/pipermail/ntg-context/2012/066676.html
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# [NTG-context] Bug with linenumbering
Thu May 3 20:59:32 CEST 2012
On Thu, 3 May 2012, Wolfgang Schuster wrote:
>
> Am 03.05.2012 um 17:18 schrieb Aditya Mahajan:
>
>> On Thu, 3 May 2012, Wolfgang Schuster wrote:
>>
>>> You assume \definelinenumbering has a second argument for the settings but this isn’t the case
>>
>> D'oh.
>>
>>> and you need \setuplinenumbering to set them.
>>
>> I also needed to change \setvalue to \setevalue in the example, and add an \expanded to get it to work.
>
> In MkIV \setuevalue is better the \setevalue and \normalunexpanded is faster than \expanded.
>
> The problem in your approach is that you can only set the style and color value with \definewhatever
> where I think it’s better to use \setupwhatever.
Thanks for both the tips!
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2014-09-02 06:47:58
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http://discuss.tlapl.us/msg03913.html
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# Re: [tlaplus] Understanding Diameter in TLC
I synced up with Markus Kuppe offline. Thanks Markus for explaining the issue here.
TLCGet("level") is not same as steps that we have reached in state space exploration.
Diameter=2 is the effect of how TLC explores the state space (it doesn't "execute" the spec), which causes the spec to have a diameter of two. TLC explores the state graph breadth first and not by generating each behavior individually
The Init predicate defines the set of initial states to be all states except those where Fast # Slow. In other words, the states excluded by Init are those where Done = TRUE. Once TLC has generate the initial states and evaluates the next-state relation, it only generate states with Done = TRUE. We can see this is you visualize the state space. There is no state with Done = TRUE that is not a successor of an initial state.
See image 1.
If we change Init s.t it only defines a particular states, i.e. F=3/\S=1 for N=4, TLC generates the state graph below that has a higher diameter. See image 2
On Friday, 20 November 2020 at 01:37:08 UTC-8 thisismy...@xxxxxxxxx wrote:
This problem is interesting because TLCGet("level") < 3 assertion never fails though I can use TLC to generate more than 3 error trace steps.
On Thu 19 Nov, 2020, 3:09 PM thisismy...@xxxxxxxxx <thisismy...@xxxxxxxxx wrote:
Hi
Can someone explain why is diameter of model state space is 2 with below spec ?
even though i give constant N = 50
EXTENDS Integers,
TLC \* for debugging
CONSTANTS N
VARIABLES Fast, Slow, Done
TypeOK ==
/\ Fast \in 1..N
/\ Slow \in 1..N
/\ Done \in {TRUE, FALSE}
Init ==
/\ Fast \in 1..N
/\ Slow \in 1..N
/\ Done = FALSE
\* different starting position
/\ Fast # Slow
MovePointers ==
/\ Done = FALSE
/\ Fast' = (Fast + 2) % N
/\ Slow' = (Slow + 1) % N
/\ Done' = (Fast' = Slow')
/\ UNCHANGED <<N>>
Next ==
MovePointers
\* If we are done, hare = tortoise
DetectCycle ==
IF Done = TRUE
THEN Fast = Slow \* make it # to see cycle detected
ELSE Fast # Slow
\* Failure of this invariant shows TLC ran it for cycle of size 42
RunsFor42 ==
IF Done = TRUE
THEN Fast # 42
ELSE 1 = 1
\* Failure of this invariant shows TLC ran it for numbers far apart.
LongCycle ==
IF Done = FALSE
THEN Fast < Slow + 20
ELSE 1 = 1
\* stop after levels/step
-- This invariant never fails
StopTLC ==
TLCGet("level") < 3
--
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2021-03-07 23:48:24
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https://gamedev.stackexchange.com/questions/63421/ball-vs-45-degree-slope-collision-detection/63422
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Ball vs 45-degree slope collision detection
I have a simple game in which the player moves a ball around. The ball bounces off of walls. Right now I have square walls(■) implemented: I use simple bounding-box collisions to check if the player will move into a wall when updating its x or y speed and if so, I multiply that speed with -1 to make them bounce.
However, I also want to implement triangular pieces(◢◣◤◥). To bounce back I believe one can simply use:
newxspeed = -1*yspeed;
newyspeed = -1*xspeed;
However, what I am having trouble with is the collision detection: When does the player hit the diagonal?
• I highly recommend the N tutorial (part 1, part 2) on this topic. – Chris Burt-Brown Oct 13 '13 at 12:58
• Thank you very much. That tutorial actually helped me to finally understand how to solve this. – Qqwy Oct 14 '13 at 16:24
• Actually it is very hard for me to mark one of the answers as 'the' solution since all of them helped me to understand the problem but none of them fully solved it. What should I do? – Qqwy Oct 14 '13 at 16:33
First of all in order to calculate the collision detection between a sphere(circle in 2D) and a line you need to calculate the perpendicular vector between the moving ball's center and the line, in order to calculate this distance you need to do the following:
So in order to calculate d in the above figure we need to do some steps.
1. Assume your line is using the parametric equation P(t) = S + t V note that V is the line direction can be obtained by subtracting (P2 - P1).
2. From Pythagoras:
d^2 = len(Q - S)^2 - len( proj(Q - S ) )^2
Then you expand the equation to get the following, it seems a bit complicated but it's actually not.
d = sqrt( len(Q - S)^2 - len( (Q - S) dot V )^2 / V^2 )
Where Q is the circle's center and S is any point on the line. Once the distance is less than the circle/sphere radius you need to trigger collision response which is explained in the next point.
It's incorrect to always flip the x or y component to bounce the ball, what you need to do is to reflect the velocity vector, in order to do so you need to calculate the Normal vector of the surface and use that normal for calculating the reflection vector using the following equation
R = 2 * (V dot N)* N - V
where R is the reflection vector, N is the normal of the surface and V is the Velocity vector.
In case of 45 deg your surface normal will be N = (1,1,0) with varying sign depending on which direction the normal faces (position or negative).
• You use a great equation. However, it is very hard to follow for someone who is new to vectors. It would be helpful if you split up your equation in smaller steps. – Qqwy Oct 14 '13 at 16:26
• By the way, is N a 2-dimensional or a 3-dimensional vector? Where does the third value (0) come from? – Qqwy Oct 14 '13 at 16:37
• I used 3D vectors because I assumed you are using a 3D API (and I might be wrong) in case it was true you need to set the 3rd component as 0, but anyway the equations should work for both 2D and 3D (and probably higher dimensions but that doesn't really matter). regarding the equations I can explain more but I need some time editing the answer. – concept3d Oct 14 '13 at 17:49
• I edited the answer I hope it makes more sense now.. btw I hope stackexchange can provide a convenient way to write mathematical formulas because it's pain and error prone right now. – concept3d Oct 14 '13 at 18:30
You want to measure the distance between the center of your ball and the wall, so:
solving the system that you see in the picture will give you the coordinates of point d.
Then, if the distance between point d and c is smaller or equal to the radius of the ball r, there is a collision between the ball and the wall
Balls are actually rather simple objects for collision detection. They collide with terrain when the distance between the ball's center and the edge of the terrain becomes less than the ball's radius. The position of the center of the ball should be trivial to obtain. Finding the nearest point of terrain is generally more complicated and the best way to do it depends on how the terrain is represented.
Your algorithm to calculate the new velocity after bouncing off a diagonal slope is incorrect. Inverting both the x- and y-coordinates will make the ball go back in the same direction it approached the slope from. This is fine if the ball comes at the terrain from a right angle, but fails for other angles. You'll want to negate only the component normal to the surface, e.g. when bouncing off the ceiling, you negate y, not x.
• Although this answer does not directly approach the problem, +1 for telling me about the way collision resolution should be done in this case. – Qqwy Oct 14 '13 at 16:36
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2021-01-20 23:06:26
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https://pos.sissa.it/414/453/
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Volume 414 - 41st International Conference on High Energy physics (ICHEP2022) - Heavy Ions
J/$\psi$ photoproduction and the production of dileptons via photon-photon interactions in hadronic Pb-Pb collisions measured with ALICE
R. Bailhache
Full text: Not available
How to cite
Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.
Open Access
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2022-10-03 08:28:54
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http://www.hsl.rl.ac.uk/catalogue/mi12.html
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## Version 1.1.0
This routine finds an approximate inverse $M$ of an $n×n$ sparse unsymmetric matrix $A$ by attempting to minimize the difference between $AM$ and the identity matrix in the Frobenius norm. The process may be improved by first performing a block triangularization of $A$ and then finding approximate inverses to the resulting diagonal blocks.
$y=Mz\phantom{\rule{2em}{0ex}}and\phantom{\rule{2em}{0ex}}y={M}^{T}z.$
The principal use of such an approximate inverse is likely to be in preconditioning iterative methods for solving the linear system $Ax=b$.
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2019-08-18 17:22:42
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http://math.stackexchange.com/questions/157110/simplification-of-the-expected-value-via-cdf-does-it-work-for-all-probability-d
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Simplification of the Expected Value via CDF: Does it work for ALL Probability Distributions?
If a random variable $X$ has a density $f$, then the expected value can be simplified: $$\mathbb{E}[X]=a+∫_{a}^{b}(1-F(x))dx,$$
where $F$ is the cumulative distribution function, $F(x)=\Pr(X≤x)$.
My question is: Does this simplification should work for all probability distributions, even for those that are not absolutely continuous with respect to Lebesgue-measure on $[a,b]$?
If $X$ is any real-valued random variable with support $[a,b]$ and $F(x)=\Pr(x≤X)$, is it always true that $$\mathbb{E}[X]=a+∫_{a}^{b}(1-F(x))dx$$
The answer to this question would be simple if one could generally extend integration-by-parts to Lebesgue-Stieltjes integration. However, this is not possible; see Wikipedia.
-
It is better to write capital $X$ for the random variable and lower-case $x$ for the argument to the c.d.f. or the p.d.f., instead of using the same thing for both usages. Among other advantages, you avoid the awkwardness of such expressions as "$\mathbb{P}(x \le x\prime)$", instead writing $\Pr(X\le x)$. (And notice that the code for $\Pr$ is \Pr.) – Michael Hardy Jun 12 '12 at 0:07
It does work for all Borel-measurable distributions that are supported on some subset of $[a,\infty)$, where $a$ is finite. I don't think you can do it if the support has no finite lower bound. – Michael Hardy Jun 12 '12 at 0:10
Let's use $X$ for the random variable, keeping $x$ for the variable of integration. In general, we have a probability measure $\mu$ on $I = [a,b]$ and
\eqalign{E[X] &= \int_I x\ d\mu(x) = a + \int_I (x-a) d\mu(x)\cr &= a + \int_I \int_a^x 1 \ dt \ d\mu(x) = a + \int_a^b \int_{[t,b]} 1 \ d\mu(x)\ dt \cr &= a + \int_a^b (1 - F(t)) \ dt\cr}
(note that $\int_{[t,b]} 1 \ d\mu(x) = 1 - F(t-)$, but that is $1 - F(t)$ (Lebesgue) almost everywhere)
-
This comment moves me to add another answer: Riemann--Stieltjes integration is enough for this. Look it up and/or make it an exercise to prove it. I.e. suppose you have a probability distribution on the set of Borel-measurable subsets of $\mathbb{R}$. Let $X$ be a random variable so distributed and let $F$ be the c.d.f., so $F(x)=\Pr(X\le x)$. Then $$\mathbb{E}(X) = \int_{-\infty}^\infty x \, dF(x),$$ where that is a Riemann--Stieltjes integral. If the support of the distribution has a finite a lower bound, then applying integration by parts to the Riemann--Stieltjes integral does it.
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2014-03-11 20:47:56
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https://www.physicsforums.com/threads/roots-of-a-complex-equation.301286/
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Roots of a complex equation
1. Mar 20, 2009
1. The problem statement, all variables and given/known data
Find all complex solutions to the following equation:
$$3(x^2 + y^2) + (x - iy)^2 + 2(x + iy) = 0$$
2. Relevant equations
I want to use the quadratic formula, but not sure if it applies here.
3. The attempt at a solution
This is as far as I can get. What I would like is some idea as to what technique to solve this.
2. Mar 20, 2009
praharmitra
I could give you two ways of solving this problem. I ask u the following questions first..
1. What is the solution to x + iy = 0, or say x+1 + i(y-1) = 0??? If you have answered these correctly, then I guess you should go ahead and convert your problem to a
f(x, y) + i g(x,y) = 0 type, and solve it...
2. Write z = x + iy. Thats one equation. Conjugate it. Thats another equation. Can u solve them together now??
3. Mar 21, 2009
I dont understand this.
Are you suggesting that the complex conjugates are roots?
The equation I supplied was expanded where x + iy was originally z and x - iy was "zed bar".
Does anyone know a procedure to follow to solve this?
4. Mar 21, 2009
praharmitra
k, i'll simplify it slightly more for you...
For method 1.
I want you to use the formula (a+b)^2 and open up every bracket... then club together all real terms together, and all imaginary terms (with an i) together...
So, now do u get something like f(x,y) + i g(x,y) = 0 ??? Can you tell me how to solve such a problem??
If no, then I'll just give one hint.... (x+1) + i ( y-1) = 0 is of the above form with f(x,y) = x+1 and g(x,y) = y-1.
The only solution to the above is (-1,1). can u tell me why?
For method 2.
ok, this i will solve slightly so that i can explain clearly....
put x + iy = z. then x - iy = z*
The equation becomes....
3 z z* + (z*)^2 + 2 z = 0
Now conjugate the above equation. we get
3 z* z + z^2 + 2 z* = 0
You now have two equations with variables z and z*. Solve for them..
Hint - Subtract them
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2017-12-18 09:49:14
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https://rjlipton.wordpress.com/2013/05/06/a-most-perplexing-mystery/
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The discrete log and the factoring problem
Antoine Joux is a crypto expert at Versailles Saint-Quentin-en-Yvelines University. He is also one of the crypto experts at CryptoExperts, having joined this startup company last November. His work is featured in all three of the company’s current top news items, though the asymptotic breakthrough on the exponent of finding discrete logs in small-characteristic fields which we covered last month is not among them. In its place are concrete results on two fields of medium characteristic (between a million and a billion) whose elements have bit-size 1,175 and 1,425. The news release on this concludes (emphasis in original):
[We] recommend to all cryptographic users to stop using medium prime fields.
Today I want to talk about a mystery, which I find the most puzzling problem in all of complexity theory, but which Ken thinks is “only” a sign of youth of the field.
Not the ${\mathsf{P}=\mathsf{NP}}$ question, not what is the power of quantum computers, not graph isomorphism, nor any other number of great puzzles. The single strangest problem, in my opinion, is the relationship between the discrete log problem and the integer factoring problem.
History shows that every improvement to one of the these problems seems to yield rather quickly a corresponding improvement to the other. This works for quantum and classical algorithms alike—recall that Peter Shor’s famous paper solved both problems at one stroke. As related here:
Historically, it has been the case that an algorithmic advance in either problem, factoring or discrete logs, was then applied to the other. This suggests that the degrees of difficulty of both problems are closely linked, and that any breakthrough, either positive or negative, will affect both problems equally.
## The Problems
Let me restate the two problems for comparison, where ${p,q}$ are primes.
• The discrete log problem is: Given ${y=a^{x} \bmod p}$, find the value of ${x}$.
• The factoring problem is: Given ${y=p\cdot q}$, find the value of ${p,q}$.
What possible relationship is there between these problems? One concerns the structure of the finite field modulo some large prime ${p}$. The other is about the multiplicative structure of the integers.
There is no known reduction from one to the other, at least none known to me. This is the great mystery.
However, they have a common parent. Consider the finite ring of integers relatively prime to ${y = pq}$, with operations modulo ${y}$. Every element ${a}$ has an order ${x}$, meaning a least integer ${x > 0}$ giving ${a^x = 1}$ (mod ${y}$), and this is the period of the powers of ${m}$. In essence, ${x}$ is the discrete logarithm of ${1}$ in base ${a}$ in this ring. Shor’s algorithm gives a high-enough-to-amplify probability that a randomly chosen ${a}$ has a findable period ${x}$ that helps produce a factor of ${y}$.
That the same idea works for discrete logarithms in fields indicates that this common parent problem captures at least some techniques that work for both factoring and discrete log. This MathOverflow item points out the importance of this being in turn a case of the hidden-subgroup problem for Abelian groups. Ken feels that only recently is the area finding techniques that really separate the problems from their parent, an indication of the field shedding its youth. This paper by Joux is an example, as we explain next.
## The Breakthrough
The new wrinkle—a sign of maturity—is that the improvement in the running time depends on the characteristic of the field, and is greatest when it is fixed. Recall that in a finite field ${\mathbb{F}_{p^k}}$, ${p}$ is the characteristic and ${k}$ is the degree of extension from the prime field ${\mathbb{F}_p}$. The prime field is the same as the field ${\mathbb{Z}_p}$ of integers modulo ${p}$, but ${\mathbb{F}_{p^k}}$ should not be confused with the integers mod ${p^k}$, which do not form a field. The idea of characteristic is also a distinction from ${\mathbb{Z}_{pq}}$.
Running times of algorithms for factoring and discrete log have heretofore been expressed in terms of the “${L}$-function,” whose general form is defined by
$\displaystyle L_Q(\beta,c) = \exp((c+o(1))(\log Q)^\beta(\log\log Q)^{1-\beta}).$
Here we may have ${Q = p^k}$. The length of the input is essentially ${n = O(\log Q) = O(k\log p)}$. The backbone of this formula is ${\exp(cn^\beta)}$, while the ${(\log\log Q)^{1-\beta}}$ factor acts as a counterweight. The main focus is the exponent ${\beta}$. If taken only thus far, the characteristic ${p}$ seems subservient to the degree ${k}$ in the time, since the formula takes the logarithm of ${p}$.
The relation between ${c}$ and ${\beta}$ partly depends on how two phases of the algorithms are balanced, a “sieving phase” in which information about specific field elements is compiled, and a “linear algebra phase” for the main computation. Once other parameters are chosen to effect the balance, the ${c}$ part is often bounded and ignorable, yielding the simpler designation ${L(\beta)}$ for the formula. The game plan is to improve the axis of the ${c}$-vs.-${\beta}$ tradeoff, to get lower ${\beta}$.
The breakthrough by Joux is to do this when ${p}$ is bounded. This is important while bootstrapping the algorithm through extension fields ${\mathbb{F}_{q^k}}$ and ${\mathbb{F}_{q^{2k}}}$, where ${q}$ itself is a power ${p^\ell}$ making ${p^\ell \approx k}$. The result is:
Theorem: For bounded ${p}$, discrete logarithms in fields of characteristic ${p}$ can be computed in time ${L(1/4 + o(1))}$.
His paper itself emphasizes two new ideas, which we express in terms we’ve described earlier on this blog.
## Ingredients
Many “sieving” algorithms for factoring and discrete log begin with what strikes us as the opposite idea to the famous Sieve of Eratosthenes. That sieve works to find primes by crossing off numbers with factors. The newer sieves have the opposite goal: they want to find integers with lots of small factors.
A big reason lots of small factors are useful is that they help build Chinese Remainder representations for large numbers while keeping their own arithmetic simple. More generally put, the small factors help generate a lot of congruences—and the large numbers with those factors serve as common zero-points for those congruences. As a general strategy point, the more small factors with known multiples, the better. Here is where we mention one idea from our old post that is already presumed by Joux:
1. Integers ${\longrightarrow}$ polynomials: As we wrote before, integers share many properties with polynomials over the integers, but polynomials are usually much easier to handle. For example, the analogue of the famous Goldbach conjecture is solved for polynomials but is still open for integers: Every nontrivial polynomial over the integers can be written as a sum of two irreducible polynomials. For another example, the deterministic primality algorithm of Manindra Agrawal, Neeraj Kayal, and Nitin Saxena begins by introducing polynomials over one variable.
In Joux’s case, polynomials are used to construct extensions of finite fields to begin with. Joux operates further on these extensing polynomials. Adding one or a few variables creates more ways to define small factors. Thus using polynomials already constitutes a way to amplify, but Joux found a new way to carry this further.
2. Amplifying Linearity: If ${g(x)}$ is a polynomial known to split into many small factors, then for any field constant ${a}$, the linearly transformed polynomial ${g(ax)}$ also has this property. This was previously known and exploited as far as it can go. Joux noticed that certain ostensibly-nonlinear transformations could wind up having the same effect. Well, a transformation of the form
$\displaystyle h(x) = \frac{ax+b}{cx+d}.$
is sometimes called a “fractional linear transformation,” and in complex analysis is known as a Möbius transformation. Now ${g(h(x))}$ is not a polynomial, but it becomes a polynomial ${g'}$ again upon multiplication by ${(cx+d)^D}$ where ${D}$ is the degree of ${g}$.
Joux proves that if you take a polynomial ${g}$ that splits into small factors over the base field ${\mathbb{F}_q}$, and take any ${a,b,c,d}$ in an extension field ${\mathbb{F}_{q^k}}$ such that ${ad \neq bc}$, then the resulting polynomial ${g'}$ over ${\mathbb{F}_{q^k}}$ likewise splits into small factors over ${\mathbb{F}_{q^k}}$ (though irreducibility and degrees may not be preserved from the corresponding factors of ${g}$).
The amplifying advantage is that whereas the ${g(ax)}$ transformation gives at most the size of the field number of new polynomials, the ${g'}$ transform gives approximately the cube of that size—after eliminating redundancies and trivialities that happen when ${a,b,c,d}$ are all in the base field ${\mathbb{F}_q}$, for example. He also employs transforms by ratios of two quadratic polynomials.
3. Seeding More than Sieving: The cubic advantage from the last step is so great that the algorithm can dispense with sieving steps needed to build a base of small factors via search. Instead we can “seed” the base with some polynomials known to have many small factors in advance. Joux starts with the simple case of ${x^q - x}$ splitting into linear factors over ${\mathbb{F}_q}$. A further fact of particular importance is:
Lemma: A polynomial ${x^{n+1} + a x^{n} + b x + c}$ splits (into linear factors) over ${\mathbb{F}_p}$ if and only if
$\displaystyle x^{(n+1)} + x^{n} + b a^{-n} x + c a^{-n-1}$
also splits.
## The Future
When I saw Dan Boneh at the RSA conference, as I first related here, he felt very confident that Joux’s work would lead soon an improvement in factoring. We will see if these happens, but his feeling needs to be taken quite seriously since he is an expert in computational number theory. I mentioned this to Lance Fortnow the other day and he immediately offered to make a bet with me. We have yet to work out the details, but I believe we will firm up the bet shortly.
## Open Problems
Would you like to take sides on the bet? Will factoring get improved too? What about the mystery? Is there even some heuristic that explains why these problems are related, in a more detailed way than their common parentage?
1. May 6, 2013 7:50 pm
Just a thought, more loose than lucid most likely —
There is another kind of “discrete logarithm” that I used to call the “vector logarithm” of a positive integer $n.$ Consider the primes factorization of $n$ and write the exponents of the primes in order as a coordinate tuple $(e_1, \ldots, e_k, 0, 0, 0, \ldots),$ where $e_j = 0$ for any prime $p_j$ not dividing $n$ and where the exponents are all $0$ after some finite point. Then multiplying two positive integers maps to adding the corresponding vectors.
2. May 7, 2013 1:16 pm
have always wondered about this vaguely myself somewhat, the discrete log/factoring link (but of course not on this same level of sophistication!) wondering if there is some [complexity theory?] problem equivalent yet less complex than the hidden subgroup problem that is rather involved to state, and moreover while there are remarkable/at other times exotic individual cases [eg barringtons thm, or the symmetric group] — its never seemed to me that group theory is highly central or fundamental to TCS. this seems somewhat surprising, given its paramount importance in theoretical math, that there is not some deeper link. maybe a missing important bridge thm? have long suspected there could maybe be a link to automated theorem proving, eg/ie maybe a group structure in automated proofs, etc., or maybe some key link to certain types of languages/grammars, etc.
• May 7, 2013 9:08 pm
Groups are inversely related to complexity. Any time you have symmetry in a problem space it reduces the complexity of the problem — finding one solution in an orbit gives you all the rest by way of relatively simple transformations.
May 8, 2013 2:54 pm
And yet nobody seems to work on or care to read an application of group theory to NP=?P issue.
• May 8, 2013 11:02 am
By the way, the inverse relationship between symmetry and diversity — that we see, for example, in the lattice-inverting map of a Galois correspondence — is a variation on an old theme of logic called the “inverse proportionality of comprehension and extension”.
C.S. Peirce, in his probings of the “laws of information”, found this principle to be a special case of a more general formula, saying that the reciprocal relation holds only when the quantity of information is constant.
• May 13, 2013 2:00 pm
Here are my ongoing if slightly sporadic notes on Peirce’s information theory and its relation to the logic of science —
3. May 7, 2013 7:25 pm
In the post “Factoring Could be Easy” of this blog an idea is described, which could more or less be stated as:
If n! mod m could be calculated in polynomial time for integers n,m then Factoring is in P.
Isn’t this a nice example of a technique which would work essentially for both factoring and discrete log?
As “equally” a fast way of testing if
(a^1 – y)*(a^2 – y)*…*(a^n – y) = 0 mod p
would let you calculate y = a^x mod p efficiently.
4. May 7, 2013 10:09 pm
dude… huh?! wow!! thx for the reaction/musing, but…. have no idea where you got this idea. any refs? it strikes me unfortunately, as not “not even wrong” but as spectacularly wrong! there is an extraordinary, staggering inherent/fundamental complexity in group theory eg given that the classification of the finite simple groups took something like ~1Century, dozens of mathematicians and 1000s of pages! also the monster group is an extraordinary object surely with some deeper meaning/link in TCS…. theres gotta be something to all this…. but it would seem in TCS we’re still “flying blind” in a big way….
• May 8, 2013 12:33 pm
oops fyi threaded this wrong! its in reply to this comment asserting “Groups are inversely related to complexity.”….
• May 8, 2013 1:28 pm
Well, I do love a good spectacle, but here I am just using the word complexity to describe what we might otherwise call a measure of diversity or variety in a problem space, the extent to which we have to treat problem instances as isolated cases without any class or rule that relates them to others.
May 8, 2013 2:57 pm
” … Lance Fortnow the other day and he immediately offered to make a bet …”.
I wonder how much of bet Lance would be willing to make on NP[FP] = #P.
May 9, 2013 2:49 am
If $n!$ in $log^{c}(n)$ is to factoring $N: N > n > \sqrt{n}$, then $X$ is discrete logarithm over $\mathbb{F}_{p}: p$ satisfies some property of $X$. What is $X$?
May 9, 2013 7:54 pm
Logarithms were invented to simplify multiplications, so it seems just natural that they be also involved in factoring, the inverse of multiplication.
However, I’d be interested to know if the aforementioned bet went as far as the polynomiality of factoring…
May 12, 2013 10:34 pm
In ‘Applied Cryptography’ p.262, Bruce Schneier wrote: “Computing discrete logarithms is closely related to factoring. If you can solve the discrete logarithm problem, then you can factor. (The converse has never been proven to be true.)”
May 16, 2013 8:28 am
The base is a generator. So the log is the totient of the modulus. Knowing that the latter is the product of two primes p and q, the log will then be (p-1)(q-1). So yes, you can factor with the discrete log. Got this solution while waking up. In fact, I’m typing this from my bed in California. I hope it’s not the kind of nonsense I can come up with in the morning. At least, I’ll have the feeling of having solved a mystery for a couple of hours. Not sure anyone reads these old threads anyway.
May 16, 2013 8:31 am
I mean, the log of 1.
May 16, 2013 9:29 am
Well, that’s if we’re given the generator. It is supplied in the discrete log problem, but not in the factoring problem. So it all depends on the complexity of finding a generator for the modulus’ multiplicative group. I have no idea if it’s in P. I guess I’ll have to read Schneier’s bibliography.
May 16, 2013 3:26 pm
Well it would be nice if we had uniqueness of the log of x. We don’t, but I guess the solution Schneier alluded to could be along these lines. What’s clear to me however is that there’s no abyssmal disconnect between the two problems; far from it.
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2018-03-20 11:51:45
|
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https://www.jiskha.com/questions/1431957/a-spherical-balloon-is-being-inflated-let-r-t-3t-can-represents-its-radius-at-time-t
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# Calculus
A spherical balloon is being inflated.Let r(t)=3t can represents its radius at time t seconds and let g(r)=4/3 pie r be the volume of the same balloon if its radius is r.Write (g.r) in terms of t and describe what it represents.
1. 👍 0
2. 👎 0
3. 👁 284
1. that's PI, not pie!!
and I have no idea what g.r means.
g(r) = 4/3 pi r^3 = 4/3 pi (3t)^3
If you want, you can expand that out. This is just Algebra I, right?
1. 👍 0
2. 👎 0
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2020-11-24 06:50:13
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https://crypto.stackexchange.com/questions/33298/key-derivation-using-the-main-aes-encryption-key-and-some-plain-text
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# Key derivation using the main AES encryption key and some plain text?
I am using AES 128 bit encryption in java. I have an encryption key and 10 digit plain text.
I want to derive a key for HMAC using the AES encryption key and the 10 digit plain text. This plain text will remain fixed in my case.
How can I derive key for HMAC using the above ingredients.
To be more precise please refer below diagram:- How can I achieve to get a new key from the respective input.
• This is quite confusing. Why do you want a new key for HMAC? Also, there are many ways to "get a new key" from two inputs. You can use a key derivation function for example. Mar 2 '16 at 19:38
• HMAC has two inputs, so you use your existing AES key as one input and the plaintext as the other. Perhaps there is something that I don't understand (or it is too easy). The question is, what do you want to do with the New Key that you get out of? Also, the AES encryption key is usually the product of HMAC, not its input Mar 2 '16 at 19:39
• @d1str0 My requirement is that I need the existing AES key for encryption and the new key for signing the message before encrypting it. If there are other ways by which I can generate the key using the these two inputs then please let me know. I am using these two inputs because on the other end I want to recover the same key with the same inputs. Mar 2 '16 at 19:42
• Why are you signing before encryption? You should always encrypt then Mac. Mar 2 '16 at 19:43
• Ok, but still I need key for HMAC ? Mar 2 '16 at 19:46
NIST has defined some key derivation functions, which can be used to derive new key using one of NIST recommended MAC algorithms, AES-CMAC or HMAC. These functions can meet your needs.
The functions are defined in NIST SP 800-108: Recommendation for Key DerivationUsing Pseudorandom Functions.
Because you intent to use AES Encryption Key for encryption using AES algorithm, you need to pick key derivation mechanism that does not use AES algorithm in ways which may conflict with your use of the key. To do this, you may choose to use the functions with HMAC algorithm, for instance with SHA-256 hash function. (Do not choose AES-CMAC.)
When using construct (Counter Mode KDF, Feedback mode KDF or Double-Pipeline KDF) from NIST SP 800-108, the plaintext can be e.g. Label and the AES encryption key needs to be the KI. For instance:
Use KDF in Counter Mode, with
• h = the length of HMAC key desired
• r = 1
• KI = AES encryption key
• Label = the 10 digit plain text
• Context = empty string or suitable identifier identifying your protocol or key use purpose.
BTW, instead of deriving HMAC key from the AES encryption key, it is more common to use key derivation key to derive all required key material, i.e., to use KDK (Key Derivation Key), which derives enough key material for both the Encryption Key and the HMAC Key. This practice is recommendable because then encryption key is not used for two different purposes: encryption and key derivation. Use of keys for multiple purposes is generally considered back practice as it increases ways possible attacker may try to get information about the key.
I.e. I would recommend you to consider this instead
AES-CTR-KDK(KDK, plaintext-key) = AES-key (first 128-bits) || HMAC-key (last 256-bits).
[These are typical key sizes, customize the key sizes according to your needs.]
Furthermore, when proposing KBKDF (key-based key derivation function) I'm assuming your use of 10 digit plaintext is not used as PIN for authentication purposes. Then be careful, as it might be possible for attacker to brute-force the key space of $10^{10}$.
• Thx for your answer. I will try to implement this. Mar 2 '16 at 22:11
• I think this answer explains all points very well, but In my opinion your text (10 digit) should be introduced in Context. This point is explained in 7.5 and 7.6 of NIST SP 800-108: Recommendation for Key Derivation Using Pseudorandom Functions. Jan 1 at 10:22
I recommend HKDF - it's a key-based key derivation function (as opposed to password-based). It accepts an already-high entropy input (your AES key), a salt (you could use your "plain text"), and optional info, for contextualizing the key. It would be better to use your main key to derive two separate keys, one for encryption, and one for HMAC... and using the info parameter to distinguish the two keys.
In this case, this answer also should be a comment (I'm sorry!). @Hunter, the NIST SP 800-108 define the use of hmac as PRF. For example, in python, you can use:
hash_algo = hashes.SHA512()
kdf = KBKDFHMAC(algorithm=hash_algo,
mode=Mode.CounterMode,
length=32,
rlen=4,
llen=4,
location=CounterLocation.BeforeFixed, label=label,
context=context,
fixed=None,
backend=backend)
derived_key = kdf.derive(keyMaster)
As you can view, you have a option the select the algorithm hash. The package used is cryptography.hazmat.primitives.kdf.kbkdf, that includes KBKDFHMAC.
Regards
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2021-09-27 23:26:51
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https://math.stackexchange.com/questions/4006878/prove-sqrtx-124x-1x-25x-222x-16x-25-is-convex
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# Prove $\sqrt{x_1^2+4x_1x_2+5x_2^2+2x_1+6x_2+5}$ is convex
I came across a question which contained this $$\sqrt{x_1^2+4x_1x_2+5x_2^2+2x_1+6x_2+5}$$
I needed to prove $$x_1^2+4x_1x_2+5x_2^2+2x_1+6x_2+5\geq0$$ which I've done by just showing that the minimum is 3 and it's convex function.
The second part asked to show that the whole problem is convex but I didn't know how to prove $$\sqrt{x_1^2+4x_1x_2+5x_2^2+2x_1+6x_2+5}$$ is convex which is same as asking $$\sqrt{x^TAx+b^Tx+c}$$ is convex where $$x^TAx+b^Tx+c\geq0$$. any help?
Extended hint: a non-negative quadratic function with positive-definite $$A$$ can be written as $$x^TAx+b^Tx+c=\|M(x-x_0)\|^2+d^2.$$ It can be done by completing the square in $$x$$ or simply identifying $$M$$, $$x_0$$ and $$d$$ from the equation: $$A=M^TM$$, $$d^2$$ is the minimum value ($$3$$ in your case) and $$x_0$$ is the minimizer.
Now the function you are dealing with is $$f(x)=\left\|\begin{matrix}M(x-x_0)\cr d\end{matrix}\right\|.$$ It could be proven to be convex directly by definition using triangle inequality (or as a composition of the convex norm-function and the affine one).
• cool cool, you use $(x-x_0)A(x-x_0) + d^2 = x^TAx + b^Tx + c$, which is true since $x_0$ satisfies $2Ax+b=0$ Feb 1 at 2:37
Note that $$x^\top A x + b^\top x + c = \begin{bmatrix}x^\top&1\end{bmatrix}\begin{bmatrix}A&b/2\\b^\top/2 & c\end{bmatrix}\begin{bmatrix}x\\1\end{bmatrix} = \tilde{x}^\top\tilde{A}\tilde{x}.$$
Now, we test $$\tilde{A}$$ for positive definiteness by calculating its principal minors (I'll skip most of the calculations). We have $$A = \begin{bmatrix}1&2\\2&5\end{bmatrix},\quad b=\begin{bmatrix}2\\6\end{bmatrix},\quad c = 5,$$ thus, the first principal minor is $$1$$, and we have $$\det(A)>0$$, which is the second principal minor of $$\tilde{A}$$. All that remains is to test $$\det{\tilde{A}}$$, which is (using Schur): $$c\det(A - \frac{bb^\top}{4c}) = 5\begin{vmatrix}1 - 1/5& 2-3/5\\2-3/5 &5-9/5\end{vmatrix} = 3 > 0.$$
Since $$\tilde{A}$$ is positive definite, then $$\lVert \tilde{x} \rVert_{\tilde{A}} = \sqrt{\tilde{x}^\top \tilde{A} \tilde{x}}$$ is a norm, and thus convex.
• your test for positive definite is not complete, it should depend on $b$ Feb 1 at 3:28
• @LinAlg My mistake, you're right, I'll edit the answer. Feb 1 at 3:35
$$f(x)=\sqrt{x^TAx+b^Tx+c}$$ is convex where $$x^TAx+b^Tx+c\geq0$$
$$A$$ should be assumed positive definite (e.g., $$\sqrt{25-x^2}$$ is not convex on $$[0,5]$$).
Proof
Some nonnegative polynomials can be written as a sum of squares. By Hilbert's Theorem, that is the case for every second degree polynomial. This allows you to write the function as a norm.
Failed attempt below
For convenience I assume that $$A$$ is symmetric. The gradient of $$f$$ is $$Df=\frac{2Ax + b}{2\sqrt{x^TAx+b^Tx+c}}$$ and the Hessian is: $$Hf =\frac{4\sqrt{x^TAx+b^Tx+c}A - (2Ax+b)(2Ax+b)^T/\sqrt{x^TAx+b^Tx+c}}{4(x^TAx+b^Tx+c)}$$ Multiplying the numerator and denominator with $$\sqrt{x^TAx+b^Tx+c}$$ yields: $$Hf =\frac{4(x^TAx+b^Tx+c)A - (2Ax+b)(2Ax+b)^T}{4(x^TAx+b^Tx+c)^{3/2}}$$ The denominator is positive on the interior of the feasible region, so all we need to check is if the numerator is positive semidefinite. In the one dimensional case, the numerator simplifies to $$4ac-b^2$$, which is indeed nonnegative because $$ax^2+bx+c=0$$ has at most one solution. Looking at derivations of the abc-formula, completing the square might be the key to a solution, but I do not yet see how it can be done. Another issue in higher dimensions is that $$x^TAxA$$ does not cancel against $$Axx^TA$$. Perhaps the statement is false after all.
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2021-09-27 05:16:10
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https://nullcount.com/add-a-disk-to-lvm-volume-group-and-create-storage-volume-in-proxmox/
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# Add a Disk to LVM Volume Group and Create Storage Volume in Proxmox
To create an LVM (Logical Volume Manager) volume in Linux, you can follow these steps:
1. Open the terminal and type the following command to list the available disks:
sudo fdisk -l
1. Identify the disk that you want to use for the LVM volume, and create a new partition using the fdisk utility:
sudo fdisk /dev/sdX
Note: Replace “sdX” with the appropriate drive identifier for your disk, such as “sda” or “sdb”.
1. In the fdisk utility, type the following command to create a new partition:
n
Note: You can press “d” to delete any existing partitions beforehand
1. Follow the on-screen prompts to configure the new partition. When prompted for the partition type, select “Linux LVM”.
2. After creating the partition, type the following command to write the changes to the disk:
w
1. Create a new physical volume on the new partition using the following command:
sudo pvcreate /dev/sdXN
Note: Replace “sdXN” with the appropriate partition identifier for the new partition, such as “sda1” or “sdb2”.
Go ahead and repeat these steps for every disk you want to add to the volume group.
1. Create a new volume group using the following command, where “myvolume” is a name you choose for the new volume group.
sudo vgcreate myvolume /dev/sdXN /dev/sdYN <keep_chaining_physical volumes_you_want_to_add>
1. List your existing volume groups with:
sudo vgdisplay
You should see the group you just created.
1. Open up Proxmox Web GUI and click on “Datacenter” > “Storage” > “Add” > “LVM”. Then select the volume group, and give your new storage volume a name (a.k.a. “ID”).
Now you have created a virtual group of physical disks that Proxmox can read as a single storage volume. Any disks you add to the group will be essentially RAID0, no redundancy with data striped across disks.
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2023-04-02 02:41:15
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http://www.maths.kisogo.com/index.php?title=Vertex_scheme_of_an_abstract_simplicial_complex
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# Vertex scheme of an abstract simplicial complex
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Not important as I wont be examined but I think it is very important to the subject! See Abstract simplicial complex as it has the same note and is why this page was created
## Definition
Let [ilmath]K[/ilmath] be a simplicial complex and let [ilmath]V_K[/ilmath] be the vertex set of [ilmath]K[/ilmath] (not to be confused with the vertex set of an abstract simplicial complex), then we may define [ilmath]\mathcal{K} [/ilmath] - an abstract simplicial complex - as follows[1]:
• $\mathcal{K}:\eq\left\{\{a_0,\ldots,a_n\}\in \mathcal{P}(V_K)\ \big\vert\ \text{Span}(a_0,\ldots,a_n)\in K\right\}$Warning:[Note 1] - that is to say [ilmath]\mathcal{K} [/ilmath] is the set containing all collections of vertices such that the vertices span a simplex in [ilmath]K[/ilmath]
## Notes
1. [ilmath]n\in\mathbb{N}_0[/ilmath] here so [ilmath]n[/ilmath] may be zero, we are expressing our interest in only those finite members of [ilmath]\mathcal{P}(V_K)[/ilmath] here, and that are non-empty.
• TODO: This needs to be rewritten!
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2021-03-01 12:33:58
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http://mymathforum.com/new-users/346719-probability.html
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My Math Forum Probability
New Users Post up here and introduce yourself!
July 8th, 2019, 05:55 AM #1 Newbie Joined: Jun 2019 From: Melbourne Posts: 4 Thanks: 0 Probability If a man has 15 balls in his left pocket and 12 balls in his right pocket and repeatedly grabs either box at random to take a ball out, what is the probability that he gets to a point where he has 10 balls in each pocket?
July 8th, 2019, 01:54 PM #2 Global Moderator Joined: May 2007 Posts: 6,822 Thanks: 723 Assuming each pocket equally likely to be chosen, the question becomes: after choosing 7 balls, what is the probability that he has chosen 5 balls from the left and 2 balls from the right. The answer is the binomial term $\frac{1}{2^7}\binom{7}{5}=0.1640625$.
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Similar Threads Thread Thread Starter Forum Replies Last Post Devans99 Probability and Statistics 4 June 25th, 2018 01:06 PM rivaaa Advanced Statistics 5 November 5th, 2015 01:51 PM hbonstrom Applied Math 0 November 17th, 2012 07:11 PM token22 Advanced Statistics 2 April 26th, 2012 03:28 PM naspek Calculus 1 December 15th, 2009 01:18 PM
Contact - Home - Forums - Cryptocurrency Forum - Top
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2019-09-18 20:24:17
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https://runestone.academy/ns/books/published/TeacherCSP/CSPNameTurtles/turtleFAP.html
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# More Turtle Procedures and Functions¶
Turtles can do more than go forward, turn left, and turn right. The table below lists more turtle procedures and functions.
Name
Input
Description
backward
amount
Moves the turle backward by the specified amount
color
colorname
Sets the color for drawing. Use ‘red’, ‘black’, etc
forward
amount
Moves the turtle forward by the specified amount
goto
x,y
Moves the turtle to position x,y
left
angle
Turns the turtle counter clockwise by the specified angle
pendown
None
Puts down the turtles tail so that it draws when it moves
penup
None
Picks up the turtles tail so that it doesn’t draw when it moves
pensize
width
Sets the width of the pen for drawing
right
angle
Turns the turtle clockwise by the specified angle
angle
Turns the turtle to face the given heading. East is 0, north is 90, west is 180, and south is 270.
Turtle
None
Creates and returns a new turtle object
You can draw a block style C with a turtle. Can you draw more than one letter? You would have to use the penup() procedure to pick up the pen and then move to the new location to draw the second letter and then put the pen down using pendown() as shown in the example below.
The space that the turtle draws in is 320 by 320 pixels. The center of the space is at x=0, y=0.
The program below uses the goto(x,y) to move to the top left corner before drawing a square that nearly fills the drawing space.
Note
Remember to put the pen down again after you have picked it up if you want to draw a line!
Mixed up programs
The following program uses a turtle to draw a capital F as shown in the picture to the left of this text, <img src=”../_static/DrawFwGT.png” width=”200” align=”left” hspace=”10” vspace=”5” /> but the lines are mixed up. The program should do all necessary set-up: import the turtle module, get the space to draw on, and create the turtle. It should draw the lines in the order shown by the numbers in the picture on the left.<br /><br /><p>Drag the needed blocks of statements from the left column to the right column and put them in the right order. There may be extra blocks that are not needed in a correct solution. Then click on <i>Check Me</i> to see if you are right. You will be told if any of the lines are in the wrong order or are the wrong blocks.</p>
The following program uses a turtle to draw a capital A as shown in the picture to the left of this text, <img src=”../_static/DrawABig.png” width=”200” align=”left” hspace=”10” vspace=”5” /> but the lines are mixed up. The program should do all necessary set-up: import the turtle module, get the space to draw on, and create the turtle. It should draw the lines in the order shown by the numbers in the picture on the left. <br /><br /><p>Drag the needed blocks of statements from the left column to the right column and put them in the right order. There may be additional blocks that are not needed in a correct solution. Then click on <i>Check Me</i> to see if you are right. You will be told if any of the lines are in the wrong order or are the wrong blocks.</p>
You can change the color and pensize that you draw with as well.
Use the area below to try to draw to draw your initials in block style with two different colors.
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2023-02-01 05:49:16
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https://electronics.stackexchange.com/questions/498200/the-difference-between-these-two-d-latch-circuits/498266
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# The difference between these two D latch circuits
I simulated both ones and could not see any difference in functionality. So, what is the need for that extra NOT gate? When it is preferred?
• That's a latch, not a D flip-flop. When E or CLK is high, the data passes through. A D flip-flop has outputs that only change on a particular clock edge (rising or falling, depending on the design). – Spehro Pefhany May 7 '20 at 17:46
• @SpehroPefhany hmmm, ok, then it should be edge triggered to be a flip flop, thanks, noted. Any idea on the use/necessity of that NOT gate? – muyustan May 7 '20 at 18:01
• It is obvious the two circuits are equivalent... but more important questions are, "What is the idea behind them?", "For what purpose were they created?". Related questions are, "Is it possible to create an asynchronous D latch? Is it possible to create a synchronous RS latch? Does it make sense?" – Circuit fantasist May 8 '20 at 15:55
• @Circuitfantasist nice questions to think about – muyustan May 8 '20 at 16:27
If you look closely at the circuits, the pin the NOT gate is driving is driven from the output of the NAND gate in the other circuit. Basically, they get a NOT gate for free with the NAND gate, and they're making use of it.
When is it preferred? I would say never. If it is possible to get the same functionality without the NOT gate without affecting performance, then there is no reason for it. Additionally, removing the NOT gate decreases the load on the input, and it only results in that one NAND gate driving two pins instead of one. However, at the same time, nobody builds latches or flip flops from logic gates these days, instead more optimized, transistor-level circuits are used to increase performance and reduce area.
If you're working with 7400 series logic, you would use a 7475, 7477, or similar latch or flip-flop chip, which gives you multiple latches in one chip instead of using a whole 7400 quad NAND gate chip for one latch.
They are more-or-less equivalent but the timing is not identical, especially for data stable and runt pulses on E in the second circuit. There may be other differences you can find.
• +1: For mentioning timing. – copper.hat May 8 '20 at 2:25
They are logically equivalent. The only apparent difference is at the bottom left NAND gate. For the top circuit the output of this NAND gate is obviously $$\overline{\overline{D}\cdot CLK}$$
In the bottom circuit the output of the NAND gate is $$\overline{\overline{(D\cdot E)}\cdot E} = \overline{(\overline{D}+\overline{E})\cdot E} = \overline{\overline{D}\cdot E + \overline{E}\cdot E} = \overline{\overline{D}\cdot E}$$ which is the same logic function if you substitute $$\E\$$ for $$\CLK\$$.
• Nice demonstration on how they are equal to each other mathematically. – muyustan May 7 '20 at 18:30
Probably, the first circuit was formally synthesized by a theorist who was not interested in how it would be implemented... while the second circuit was invented by an engineer who knew about the existence of 7400... This was exactly the reason we were guided by in the 90's, when we made setups for investigating various latches and flip-flops in the laboratory on digital circuits...
• Quad nand2 chip? – muyustan May 7 '20 at 20:18
• And apparently nobody heard of the 7475/7477, because that gives you four not-quite-independent D latches in one chip – alex.forencich May 7 '20 at 21:09
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2021-06-16 14:58:50
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https://www.gamedev.net/forums/topic/601729-travel-at-a-fraction-of-light-speed-subjective-time-question/
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# Travel at a fraction of light speed. Subjective time question.
This topic is 2762 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
To keep the calculations simple, say that a traveler had a spaceship that could instantly accelerate to 1/2 of light speed (without killing him) and he started a journey to a point in space 10 light years away to finally instantly decelerate at the moment of arrival.
How many time would it pass from the point point of view of the traveler inside the spaceship?
Intuitive thinking may make one think it'd be 20 years but I know relativistic speeds don't work that way. What would be the equations to calculate that "t"?
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See this article. This is the equation you're after (explained in the article):
.
(The "gamma" in the above equation is known as the Lorentz factor.)
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For the given values it would be:
10 / ( Root( 1-( (1/2)^2 / 1) ) ) = 11,5470
11 years and a half?
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To keep the calculations simple, say that a traveler had a spaceship that could instantly accelerate to 1/2 of light speed (without killing him) and he started a journey to a point in space 10 light years away to finally instantly decelerate at the moment of arrival.
How many time would it pass from the point point of view of the traveler inside the spaceship?
Intuitive thinking may make one think it'd be 20 years but I know relativistic speeds don't work that way. What would be the equations to calculate that "t"?
Correct me if I'm wrong, but I'm pretty sure that for the the guy in the spaceship it would be twenty years.
However, see this video for a simple explanation of time dilation. Skip to about 6:40 to avoid the junk.
Hope this helps!
Also a book called 'The Elegant Universe' explains this and many other related issues very well.
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From the traveler's perspective, the calculation is simple... Time = Distance / Speed... which is 10 / 0.5 = 20 years.
The time dilation effect occurs when you compare what an external observer sees versus what the traveler sees. In this example, more than 20 years will have passed for the rest of the universe. The reason for this is that the non-moving reference frame sees the traveler's clock running slow (the actual amount is given by the Lorentz factor as stated by Emergent). So 20 years for a slow running clock means more than 20 years will have passed for the rest of us.
However, no matter what the speed of travel is, the time taken for the journey from the traveler's point of view is the simple formula : Time = Distance / Speed.
Edit: The value of the Lorentz factor for the given setup is about 1.1547, so about 23.094 years will have passed. In your calculation you used the distance (10 light years) not the time (20 years).
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Hidden
no, wait I meant 10 years traveling at speed c (which still makes no sense I realise). The resulting value is the elapsed time for an observer outside the ship then.
[font="arial, verdana, tahoma, sans-serif"][s]no, wait I meant 10 years traveling at speed c (which still makes no sense I realise). The resulting value is the elapsed time for an observer outside the ship then[/s]
No I got that right. [/font]
Not sure about the formula Emergent provided though. As the traveler approaches the speed of light, the flow of time in his local frame decreases by which the object in which he is traveling, to him, appears to be moving faster than what it really is (covering more external distance in less time). So, his perception of the time passed since he started traveling will be inferior than the time that passes for an external observer.
I double checked this in Isaac Asimov's "Extraterrestrial Civilizations" non-fiction book (Chapter "The speed of light"). He mentions that at 293,800 km/s or 98% of the speed of light, time inside a spaceship will flow at 1/5 of the rate it would if it was at rest. He states then than for a distance of 10 light years at that speed, from the point of view of an external observer (at rest in relation to the ship, say the Earth) the travel would seem to last a little longer than 10 years but for the traveler on board the ship the travel would seem to have lasted just one week.
So, again, what's the formula with which I can calculate those numbers?
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Well, if he travels at 0.5*c, then 10 lightyears means 20 years from the observers POV. But the equation should be dt'/gamma because the distance is measured in the observer's coordinate system (which is considered standing: we defined distances in Space like that).
You can verify it, because the time from the traveler's CS should be less than the time in the observers CS, because we know (from sci-fies....) if the traveler would travel with c his time would be zero. (if you substitute v = 0, dt'/gamma will be zero as expected.)
so 17.32 years is the correct answer.
Another verification:
v/c = 0.98, dt' = 1 ->
dt = 0.198997 -> 1/5, see Asimov's
EDIT: edited dt' dt confusion. Emergent's formula is perfect, it's just we are not calculating dt', but dt.
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Thanks, szecs. Could you write the formula for me in pseudo-code so I can see it better? Just pretend I'm scientific-notation blind
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[font="Courier New"]20 * ( Root( 1-( (1/2)^2 / 1) ) ) = 17.32[/font]
or
[font="Courier New"]time_for_traveler = time_for_observer * sqroot(1-(v/c)^2)[/font]
just pretending I'm not having sense of humor (which I'm really not having..)
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• ### Forum Statistics
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2018-12-19 09:53:35
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https://optimization-online.org/tag/completion-time-variance/
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## An Almost Exact Solution to the Min Completion Time Variance in a Single Machine
We consider a single machine scheduling problem to minimize the completion time variance of n jobs. This problem is known to be NP-hard and our contribution is to establish a novel bounding condition for a characterization of an optimal sequence. Specifically, we prove a necessary and sufficient condition (which can be verified in O(n\log n)) … Read more
## Sufficient condition on Schrage conjecture about the completion time variance
We consider a single machine scheduling problem to minimize the completion time variance. This roblem is known to be NP-hard. We prove that if \$p_{n-1} = p_{n-2\$, then there is an optimal solution of the form \$(n,n-2,n-3,…,n-4,n-1)\$. A new lower bound are proposed for solving the problem. The test on more than 4000 instances shows … Read more
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2022-08-17 02:30:32
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https://www.physicsforums.com/threads/reverse-engineering-relativity.67388/
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# Reverse engineering relativity?
1. Mar 15, 2005
Staff Emeritus
Is it possible to start with the Lorentz transformations - I mean the whole Poincare group SO(1,3) - and derive that there exists an invariant speed? Is this a known derivation? I suppose we could assume a "flat" manifold with metric Diag(1,-1,-1,-1,).
2. Mar 15, 2005
### marlon
I am just wondering, wouldn't it be one heck of a coincidence that you start with "such" transformations ? I don't see the point in doing so, to be honest. However, isn't this quite easy to prove ? Just apply those transformations to two reference frames...
marlon
3. Mar 15, 2005
### marlon
Besides, velocity being absolute is an inherent property of the Lorentz transformations, right ? They are constructed starting from the universal constant c and the homogenity of space...So if you start from the L transformations, you already have that property
marlon
4. Mar 15, 2005
### dextercioby
It would be interesting SA.But it would be cheating.You cannot teach that.Students might ask you:"Why SO(3,1)"??What would you say ?What would be the reasoning behind not follwing the Einstein approach,the aximatical one?
It sorta reminds of the "deriving Schrödinger equation" threads.
Daniel.
5. Mar 15, 2005
### robphy
Suggestion: Find the eigenvectors. (They should lie on a cone through the origin.)
It's probably easiest to leave off the translations.
To see that this is plausible, try the usual Lorentz Transformations in 2D Minkowski.
Edit: Furthermore, one should be able to show that if the eigenvalue corresponding to an eigenvector is not 1 or -1, then that eigenvector must have zero norm [with respect to the metric preserved by the transformation].
Last edited: Mar 15, 2005
6. Mar 17, 2005
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2018-02-18 20:53:29
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https://computergraphics.stackexchange.com/questions/8667/is-groupshared-memory-stored-in-l2-cache-of-gpu/8668
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# Is groupshared memory stored in L2 cache of GPU?
The article says that L1 cache is shared by work items in the same work group(aka. SM) and L2 cache is shared by different work groups.
In Direct3D, it seems that a thread group (which is specified by numthread(x, y, z)) is mapped to multi work groups, and groupshared could be accessed by all threads of the thread group (which belong to different work groups).
So, is groupshared the L2 cache of GPU? If it is, could I manipulate L1 cache directly in Direct3D?
There seems to be some confusion of terminology here. In Direct3D, you have threads and thread groups. "work item" and "work group" are generally encountered in OpenCL terminology, where a "work item" would be what a thread is in Direct3D and a "work group" corresponds to a Direct3D thread group. groupshared memory is memory accessible to all threads that are part of the same thread group.
That being said, groupshared memory is generally located in fast on-chip storage just like caches are. The whole point of groupshared memory is to take advantage of the knowledge that certain threads reside on the same part of the chip to allow them to communicate faster. And on some hardware (e.g., most of the recent NVIDIA architectures), groupshared memory and the L1 cache will actually use the same physical memory. However, that just means that one part of that memory is used as "normal" memory, accessed directly via addressing through some load-store-unit, while another part is used by the L1 cache to cache stuff…
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2021-05-14 14:17:10
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|
https://imcs.dvfu.ru/cats/static/problem_text-cid-1855036-pl-en.html?nosubmit=1
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## Problem A. Axis of choice ≡
Author: A. Klenin Time limit: 1 sec Input file: input.txt Memory limit: 256 Mb Output file: output.txt
### Statement
A rectangular area with sides parallel to coordinate axises is represented by coordinates of two opposite corners (x1, y1) and (x2, y2).
Your program must find the coordinate axis which is closest to the given rectangular area.
### Input file format
Input file contains four integers x1 y1 x2 y2 — coordinates of area corners.
### Output file format
Output file must contain a single string:
• X if a distance from area to X axis is less than to Y axis;
• Y if a distance from area to X axis is greater than to Y axis;
• XY if distances from area to X and Y axises are equal.
### Constraints
109 ≤ x1, y1, x2, y2 ≤ 109
### Sample tests
No. Input file (input.txt) Output file (output.txt)
1
30 40 10 20
Y
2
-5 20 -4 -20
X
3
1 1 1 1
XY
## Problem B. Increment and jump ≡
Author: A. Klenin Time limit: 1 sec Input file: input.txt Memory limit: 256 Mb Output file: output.txt
### Statement
Young programmer Vasya decided to learn microprocessor design. As a first step, he designed a processor with a single integer register with initial value 0, and two instructions:
• Increment register value by 1 (denoted by letter i).
• Jump back to the beginning of a program (denoted by letter j).
Vasya quickly noticed that this instruction set is not very useful, because any program containing even one jump instruction will loop forever.
Vasya's friend Petya suggested to modify a design so that after executing jump instruction processor would replace it with a special 'no-op' instruction which does nothing.
Now, Vasya's microprocessor could do something at least a little bit interesting. Vasya wants to write an emulator to explore the possibilities.
Your program must output a value in the register after the execution of a given program for Vasya's microprocessor.
### Input file format
The input file contains a single string consisting of letters i and j — a program.
### Output file format
Output file must contain single integer — value in the register.
### Constraints
Program length is between 1 and 105 characters.
### Sample tests
No. Input file (input.txt) Output file (output.txt)
1
i
1
2
j
0
3
ijjijiijji
17
## Problem C. Blanks ≡
Author: A. Klenin Time limit: 1 sec Input file: input.txt Memory limit: 256 Mb Output file: output.txt
### Statement
Young programmer Vasya works for a local government.
His first job is to automate filling of paper forms. A form is represented by a single string consisting of Latin letters, spaces and underscores ('_', ASCII 95). Substrings of underscores (known as fields) represent blank spaces left for an applicant to fill. Under each field the paper form has a small caption explaining to the applicant what data he should put over those underscores. For example:
My name is ________
name
In computerized form, captions are represented by a single string of spaces and Latin letters. Every caption is a single word. Number of captions is guaranteed to be equal to the number of fields. However, number of spaces between captions may be arbitrary, so captions are not necessarily located directly under corresponding fields.
You program must, given the representation of a form and a value for each field caption, put corresponding value into every form field.
When filling a field with a value, additional rules must be observed:
1. First and last character of the field must remain underscores.
2. Total field length must not change. If a value is longer than a field, it must be truncated on the right. If a value is shorter than a field, it must be aligned to the left.
### Input file format
First line of input file contains a string representing the form. Second line contains field captions. Third line contains an integer N — number of different captions. Following 2 × N lines contain N pairs of captions and corresponding values. Captions consist of Latin letters only. Values consist of Latin letters and spaces. All captions are different. Note that several fields may have the same caption.
### Output file format
Output file must contain a single line — a form with all fields filled in. It is guaranteed that values for all captions are present in the input. Note that spaces in the form are significant and must be preserved.
### Constraints
All strings have length from 1 to 1000 characters. All fields consist of at least 3 underscores. 0 ≤ N ≤ 100
### Sample tests
No. Input file (input.txt) Output file (output.txt)
1
Hello _________ welcome to our ___
name location
2
name
Vasya
location
Eastowner city
Hello _Vasya___ welcome to our _E_
2
___ ____
a a
1
a
bb
_b_ _bb_
0.039s 0.006s 11
|
2022-06-26 01:33:20
|
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|
https://math.stackexchange.com/questions/2255228/how-to-solve-this-y-y-2-sec3x-differential-equation
|
# How to solve this $y'' +y = 2 \sec(3x)$ differential equation?
I am having trouble with differential equations. $$y'' +y = 2 \sec(3x)$$
I really appreciate if you help me.
I assume you know the homogeneous solution $y'' + y = 0$ leads to $y(x) = c_1 \sin x + c_2 \cos x$. The particular solution is probably easiest to solve using variation of parameters. The guess of $\sec (3x)$ or $\tan (3x)$ does not work. Guessing is usually only done with $x^n,e^{ct},\sin, \cos$.
|
2020-01-27 09:24:29
|
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|
https://kcuilla.github.io/reactablefmtr/reference/highlight_min_max.html
|
The highlight_min_max() function assigns a font color and/or background color to both the minimum and maximum values in a column. It should be placed within the style argument in reactable::colDef.
## Usage
highlight_min_max(
data,
min_font_color = "red",
max_font_color = "green",
min_highlighter = NULL,
max_highlighter = NULL
)
## Arguments
data
Dataset containing at least one numeric column.
min_font_color
color to assign to minimum value in a column. Default color is red.
max_font_color
color to assign to maximum value in a column. Default color is green.
min_highlighter
color to assign the background of a cell containing minimum value in a column.
max_highlighter
color to assign the background of a cell containing maximum value in a column.
## Value
a function that applies a color to the minimum and maximum values in a column of numeric values.
## Examples
data <- MASS::road[11:17, ]
## By default, the minimum and maximum values are bold with a red and green font color respectively
reactable(data,
defaultColDef = colDef(
style = highlight_min_max(data)))
#> Error in x$width %||% settings$fig.width * settings$dpi: non-numeric argument to binary operator ## Assign a different font color to the min and max values reactable(data, defaultColDef = colDef( style = highlight_min_max(data, min_font_color = "orange", max_font_color = "blue"))) #> Error in x$width %||% settings$fig.width * settings$dpi: non-numeric argument to binary operator
## Highlight the background of the cell for the min and max values in each column
reactable(data,
defaultColDef = colDef(
style = highlight_min_max(data,
min_highlighter = "salmon",
max_highlighter = "skyblue")))
#> Error in x$width %||% settings$fig.width * settings\$dpi: non-numeric argument to binary operator
|
2022-06-26 13:43:47
|
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|
http://opennmt.net/OpenNMT-tf/training.html
|
# Training¶
## Monitoring¶
OpenNMT-tf uses TensorBoard to log information during the training. Simply start tensorboard by setting the active log directory, e.g.:
tensorboard --logdir="."
then open the URL displayed in the shell to monitor and visualize several data, including:
• training and evaluation loss
• training speed
• learning rate
• computation graphs
• word embeddings
• decoder sampling probability
## Replicated training¶
OpenNMT-tf training can make use of multiple GPUs with in-graph replication. In this mode, the main section of the graph is replicated over multiple devices and batches are processed in parallel. The resulting graph is equivalent to train with batches N times larger, where N is the number of used GPUs.
For example, if your machine has 4 GPUs, simply add the --num_gpus option:
onmt-main train [...] --num_gpus 4
Note that evaluation and inference will run on a single device.
## Distributed training¶
OpenNMT-tf also supports asynchronous distributed training with between-graph replication. In this mode, each graph replica processes a batch independently, compute the gradients, and asynchronously update a shared set of parameters.
To enable distributed training, the user should use the train_and_eval run type and set on the command line:
• a chief worker host that runs a training loop and manages checkpoints, summaries, etc.
• a list of worker hosts that run a training loop
• a list of parameter server hosts that synchronize the parameters
Then a training instance should be started on each host with a selected task, e.g.:
CUDA_VISIBLE_DEVICES=0 onmt-main train_and_eval [...] \
--ps_hosts localhost:2222 \
--chief_host localhost:2223 \
--worker_hosts localhost:2224,localhost:2225 \
will start the worker 1 on the current machine and first GPU. By setting CUDA_VISIBLE_DEVICES correctly, asynchronous distributed training can be run on a single multi-GPU machine.
For more details, see the documentation of tf.estimator.train_and_evaluate. Also see tensorflow/ecosystem to integrate distributed training with open-source frameworks like Docker or Kubernetes.
Note: distributed training will also split the training directory model_dir accross the instances. This could impact features that restore checkpoints like inference, manual export, or checkpoint averaging. The recommend approach to properly support these features while running distributed training is to store the model_dir on a shared filesystem, e.g. by using HDFS.
## Mixed precision training¶
Thanks to work from NVIDIA, OpenNMT-tf supports training models using FP16 computation. Mixed precision training is automatically enabled when the data type of the inputters is defined to be tf.float16. See for example the predefined model TransformerFP16:
onmt-main train [...] --model_type TransformerFP16
Additional training configurations are available to tune the loss scaling algorithm:
params:
# (optional) For mixed precision training, the loss scaling to apply (a constant value or
# an automatic scaling algorithm: "backoff", "logmax", default: "backoff")
loss_scale: backoff
# (optional) For mixed precision training, the additional parameters to pass the loss scale
# (see the source file opennmt/optimizers/mixed_precision_wrapper.py).
loss_scale_params:
scale_min: 1.0
step_factor: 2.0
For more information about the implementation and get expert recommendation on how to maximize performance, see the OpenSeq2Seq’s documentation. Currently, mixed precision training requires Volta GPUs and the NVIDIA’s TensorFlow Docker image.
If you want to convert an existing checkpoint to FP16 from FP32 (or vice-versa), see the script onmt-convert-checkpoint. Typically, it is useful when you want to train using FP16 but still release a model in FP32, e.g.:
onmt-convert-checkpoint --model_dir ende-fp16/ --output_dir ende-fp32/ --target_dtype float32
The checkpoint generated in ende-fp32/ can then be used in infer or export run types.
|
2018-10-23 12:30:48
|
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|